The current in the primary is 5.42 A (or 5420 mA) and the final answer is, (a) 1.9 V, (b) 0.088 A and (c) 1.4E-3 V.
Primary turns (Np) = 680
Secondary turns (Ns) = 11
Primary voltage (Vp) = 120 Vrms
(a) When there is no load, it means the secondary winding is an open circuit.
Therefore, the voltage across the secondary (Vs) can be calculated using the turns ratio formula as:
Vs/Vp = Ns/NpVs/120 = 11/680Vs = 1.9 V
(b) Resistive load in secondary = 22 ΩThe current in the secondary (Is) can be calculated using Ohm’s law as:Is = Vs/Rs
Where Rs = 22 Ω, Vs = 1.9 VIs = Vs/Rs = 1.9/22 = 0.088 A (or 88 mA)
(c) The current in the primary (Ip) can be calculated using the relation:
Vs/Vp = Ns/NpIs/IpIp = Is × Np/NsIp = 0.088 × 680/11Ip = 5.42 A
Therefore, the current in the primary is 5.42 A (or 5420 mA).
Hence, the final answer is, (a) 1.9 V, (b) 0.088 A and (c) 1.4E-3 V.
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A solution consisting of 30% MgSO4 and 70% H2O is cooled to 60°F. During cooling, 5% of the water evaporates.
whole system. How many kilograms of crystals will be obtained from 1000 kg of original mixture?
The amount of MgSO4 crystals obtained from the 1000 kg of original mixture is 85.5 kg given that a solution consisting of 30% MgSO4 and 70% H2O is cooled to 60°F.
The total amount of the mixture is 1000 kg. The solution consists of 30% MgSO4 and 70% H2O.The weight of MgSO4 in the initial solution = 30% of 1000 kg = 300 kg
The weight of water in the initial solution = 70% of 1000 kg = 700 kg
The mass of the solution (mixture) = 1000 kg
During cooling, 5% of water evaporates => The mass of water in the final mixture = 0.95 × 700 kg = 665 kg
The mass of MgSO4 in the final mixture = 300 kg
Remaining mixture (H2O) after evaporation = 665 kg
The amount of MgSO4 crystals obtained = Final MgSO4 weight – Initial MgSO4 weight = 300 – (1000 – 665) × 0.3 = 85.5 kg
Therefore, the amount of MgSO4 crystals obtained from the 1000 kg of original mixture is 85.5 kg.
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A car comes to a stop six seconds after the driver applies the brakes. While the brakes are on, the following velocities are recorded:
The car has a negative acceleration of 4.17 m/s². It comes to a stop after six seconds as the velocity is decreasing at a constant rate of 4.17 m/s every second.
A car comes to a stop six seconds after the driver applies the brakes.
While the brakes are on, the following velocities are recorded:
Initial velocity, u = 25 m/sFinal velocity, v = 0 m/sTime, t = 6 s
Average acceleration, a can be calculated by the equation: a = (v - u) / t.
Therefore, substituting the values gives us:a = (0 - 25) / 6 = -4.17 m/s².
Here, the minus sign indicates that the acceleration is in the opposite direction to that of the initial velocity (deceleration).
The negative acceleration means that the velocity of the car decreases.
Therefore, the car's velocity is decreasing by 4.17 m/s every second. Hence, the car will come to a stop after six seconds as given in the problem statement.
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1111.A door is 2.5m high and 1.7m wide. Its moment of inertia is 180kgm^2. What would be its angular acceleration if you push it in the middle of the door with a force of 150N perpendicular to the door? (10 pts) What torque are you applying?(10 pts)
The angular acceleration of the door is calculated as to be 0.708 rad/s² and the torque being applied is calculated as to be 127.5 Nm.
A door is 2.5m high and 1.7m wide. Its moment of inertia is 180kgm². The torque that is being applied by a force F is given asτ = Fd, where d is the distance between the point of rotation (pivot) and the point of application of force.
Here, the force is applied at the center of the door, so the torque can be written asτ = F x (1/2w), where w is the width of the door.τ = 150 N x (1/2 x 1.7 m)τ
= 127.5 Nm
The moment of inertia of the door is given as I = 180 kg m². The angular acceleration α can be calculated as the torque divided by the moment of inertia,α = τ / Iα
= 127.5 / 180α
= 0.708 rad/s²
Therefore, the angular acceleration of the door is 0.708 rad/s².
The torque being applied is 127.5 Nm.
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An infinite line charge of uniform linear charge density λ = -2.1 µC/m lies parallel to the y axis at x = -1 m. A point charge of 1.1 µC is located at x = 2.5 m, y = 3.5 m. Find the x component of the electric field at x = 3.5 m, y = 3.0 m. kN/C Enter 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts]
In the figure shown above, a butterfly net is in a uniform electric field of magnitude E = 120 N/C. The rim, a circle of radius a = 14.3 cm, is aligned perpendicular to the field.
Find the electric flux through the netting. The normal vector of the area enclosed by the rim is in the direction of the netting.
The electric flux is:
The electric flux is 7.709091380790923. The electric field due to an infinite line charge of uniform linear charge density λ is given by:
E = k * λ / x
The electric field due to an infinite line charge of uniform linear charge density λ is given by:
E = k * λ / x
where k is the Coulomb constant and x is the distance from the line charge.
The x component of the electric field at x = 3.5 m, y = 3.0 m is:
E_x = k * λ / (3.5) = -2.86 kN/C
The electric field due to the point charge is given by:
E = k * q / r^2
where q is the charge of the point charge and r is the distance from the point charge.
The x component of the electric field due to the point charge is:
E_x = k * 1.1 * 10^-6 / ((3.5)^2 - (2.5)^2) = -0.12 kN/C
The total x component of the electric field is:
E_x = -2.86 - 0.12 = -2.98 kN/C
The electric flux through the netting is:
Φ = E * A = 120 * (math.pi * (14.3 / 100)^2) = 7.709091380790923
Therefore, the electric flux is 7.709091380790923.
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A structural steel bar is loaded by an 8 kN force at point A, a 12 kN force at point B and a 6 kN force at point C, as shown in the figure below. Determine the bending moment about each of the points. Indicate whether this bending moment is acting clockwise negative or counter-clockwise positive.
Bending moment about point A: 0 kN·m, Bending moment about point B: 0 kN·m, Bending moment about point C: 0 kN·m.
Determine the bending moment about each point due to the applied forces and indicate their direction (clockwise or counterclockwise).To determine the bending moment about each point, we need to calculate the moment created by each force at that point. The bending moment is the product of the force and the perpendicular distance from the point to the line of action of the force.
Bending moment about point A:
The force at point A is 8 kN.The perpendicular distance from point A to the line of action of the force at point A is 0 (since the force is applied at point A).Therefore, the bending moment about point A is 0 kN·m.Bending moment about point B:
The force at point B is 12 kN.The perpendicular distance from point B to the line of action of the force at point B is 0 (since the force is applied at point B).Therefore, the bending moment about point B is 0 kN·m.Bending moment about point C:
The force at point C is 6 kN.The perpendicular distance from point C to the line of action of the force at point C is 0 (since the force is applied at point C).Therefore, the bending moment about point C is 0 kN·m.All the bending moments about points A, B, and C are 0 kN·m.
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Assume you charge a comb by running it through your hair and then hold the comb next to a bar magnet. Do the electric and magnetic fields produced constitute an electromagnetic wave?(a) Yes they do, necessarily.(b) Yes they do because charged particles are moving inside the bar magnet.(c) They can, but only if the electric field of the comb and the magnetic field of the magnet are perpendicular.(d) They can, but only if both the comb and the magnet are moving. (e) They can, if either the comb or the magnet or both are accelerating.
The electric and magnetic fields produced by charging a comb and holding it next to a bar magnet do not necessarily constitute an electromagnetic wave.
Option (c) is correct
They can form an electromagnetic wave, but only if the electric field of the comb and the magnetic field of the magnet are perpendicular. The movement of charged particles inside the bar magnet, as mentioned in option (b), is not directly related to the formation of an electromagnetic wave.
Additionally, options (d) and (e) are not necessary conditions for the production of an electromagnetic wave. They can form an electromagnetic wave, but only if the electric field of the comb and the magnetic field of the magnet are perpendicular.
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An apartment has a living room whose dimensions are 2.9 mx 4.1 mx 4.7 m. Assume that the air in the room is composed of 79% nitrogen (N₂) and 21% oxygen (O₂). At a temperature of 25 °C and a pressure of 1.01 x 105 Pa, what is the mass (in grams) of the air?
The mass of air in the apartment with dimensions 2.9 mx 4.1 mx 4.7 m composed of 79% nitrogen and 21% oxygen at 25°C and 1.01 x 105 Pa is 1525.6 g.
We can use the Ideal Gas Law (PV = nRT) to solve for the mass of air in the living room.
Given: P = 1.01 x 105 Pa, V = 2.9 m x 4.1 m x 4.7 m = 56.97 m³, n (moles of air) = ?, R = 8.31 J/mol K (Universal Gas Constant), T = 25°C = 25 + 273 = 298 K.
P = nRT/V = (79/100)(1.01 x 105 Pa) + (21/100)(1.01 x 105 Pa) = 1.01 x 105 Pa (since pressure is the same for both gases)
Solving for n, we get: n = PV/RT = (1.01 x 105 Pa)(56.97 m³)/(8.31 J/mol K)(298 K) = 238.17 mol
The molar mass of air is 28.97 g/mol (approximately).
Therefore, the mass of air in the living room is:
m = n x M = (238.17 mol)(28.97 g/mol) = 6907.6 g ≈ 1525.6 g (to 3 significant figures)
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A parallel-plate capacitor with empty space between its plates is fully charged by a battery. If a dielectric (with dielectric constant equal to 2) is then placed between the plates after the battery is disconnected, which one of the following statements will be true? The capacitance will increase, and the stored electrical potential energy will increase. The capacitance will decrease, and the stored electrical potential energy will increase. The capacitance will increase, and the stored electrical potential energy will decrease. The capacitance will decrease, and the stored electrical potential energy will decrease.
When a dielectric (with a dielectric constant equal to 2) is placed between the plates of a parallel-plate capacitor with empty space between its plates after the battery is disconnected, the capacitance will increase, and the stored electrical potential energy will decrease. The correct option is - The capacitance will increase, and the stored electrical potential energy will decrease.
The capacitance of the parallel-plate capacitor with the empty space between its plates is given by;
C = ε0A/d
where C is the capacitance, ε0 is the permittivity of free space (8.85 x 10⁻¹² F/m), A is the surface area of the plates of the capacitor, and d is the distance between the plates.
When a dielectric is placed between the plates of the capacitor, the permittivity of the dielectric will replace the permittivity of free space in the equation.
Since the permittivity of the dielectric is greater than the permittivity of free space, the capacitance of the capacitor will increase by a factor equal to the dielectric constant (K) of the dielectric (C = Kε0A/d).
Thus, the capacitance will increase, and the stored electrical potential energy will decrease.
An increase in the capacitance means that more charge can be stored on the capacitor, but since the battery has already been disconnected, the voltage across the capacitor remains constant.
The stored electrical potential energy is given by;
U = 1/2 QV
where U is the stored electrical potential energy, Q is the charge stored on the capacitor, and V is the voltage across the capacitor.
Since the voltage across the capacitor remains constant, the stored electrical potential energy will decrease since the capacitance has increased.
Therefore, the correct option is- The capacitance will increase, and the stored electrical potential energy will decrease.
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Person A and B both lift an object of 50 kg to a height of 2 m. It takes person A10 seconds to lift up the object but it only takes person B 1 second to do the same. (a) How much work do A and B perform? (b) Who is more powerful? Prove
(a) Person A and Person B both perform 1000 Joules of work.
(b) Person B is more powerful.
When calculating work, we use the formula: Work = Force × Distance × cos(θ), where Force is the force applied, Distance is the distance traveled, and θ is the angle between the force and the direction of motion.
In this scenario, both Person A and Person B lift the same object to the same height, so the distance traveled is the same for both individuals. The force applied is equal to the weight of the object, which is given as 50 kg.
For Person A, it took 10 seconds to lift the object, while Person B accomplished the task in just 1 second. Since work is defined as the product of force and distance, and distance is the same for both individuals, we can conclude that the person who accomplishes the task in less time performs more work.
Therefore, Person B, who lifted the object in 1 second, is more powerful than Person A.
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6) A solar cell generates a potential difference of 0.23 V when a 4902 resistor is connected across it, and a potential difference of 0.28 V when a 98092 resistor is 1 substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is 2.4cm² and the rate per unit area at which it receives energy from light is 6.0mW/cm². What is the efficiency of the cell for converting light energy to thermal energy in the 98022 external resistor?
The efficiency of the solar cell for converting light energy to thermal energy in the 98022 external resistor is 82%.
a) Calculation of Internal Resistance
In the first case, the potential difference is 0.23 V, and the resistance is 4902Ω.From Ohm's law; the current (I) = V/RI = 0.23/4902I = 0.0000469
For the internal resistance (r); r = (V/I) - Rr
= (0.23/0.0000469) - 4902
r = 4.88 - 4902
r = -4901.87
b) Calculation of emfIn the second case, the potential difference is 0.28 V, and the resistance is 98092Ω.
From Ohm's law;
the current (I) = V/R
V= IRV = 0.28/98092
I = 0.00000285
For the emf (E),
E = V + Ir
E = 0.28 + (0.00000285 × 4902)
E = 0.2926 V
c) Calculation of efficiency
From the data given, the area (A) of the cell is 2.4cm², and the rate per unit area at which it receives energy from light is 6.0mW/cm².
So the rate at which it receives energy is;
P = (6.0 × 2.4) mW
P = 14.4 mW
From the power output in b, the current I can be calculated by;
I = P/VI = 14.4/0.28
I = 51.42mA
The power generated by the solar cell is;
P1 = IV
P1 = (51.42 × 0.23) mW
P1 = 11.82 mW
The power that is wasted in the internal resistance is;
P2 = I²r
P2 = (0.05142² × 4901.87) mW
P2 = 12.60 µW
The power that is dissipated in the external resistance is;
P3 = I²R
Eficiency (η) = (P1/P) x 100%
η = (11.82/14.4) x 100%
η = 81.875 ≈ 82%T
Therefore, the efficiency of the solar cell for converting light energy to thermal energy in the 98022 external resistor is 82%.
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Assume 3 moles of a diatomic gas has an internal energy of 10
kJ. Determine the temperature of the gas.
The temperature of the gas is 1.83 x 10^5 K.
The internal-energy of a gas is directly proportional to its temperature according to the equation:
ΔU = (3/2) * n * R * ΔT
where ΔU is the change in internal energy, n is the number of moles, R is the gas constant, and ΔT is the change in temperature.
In this case, we have ΔU = 10 kJ, n = 3 moles, and we need to find ΔT. Rearranging the equation, we get:
ΔT = (2/3) * ΔU / (n * R)
Substituting the given values, we have:
ΔT = (2/3) * (10 kJ) / (3 * R)
To find the temperature, we need to convert the units of ΔT to Kelvin. Since 1 kJ = 1000 J and the gas constant R = 8.314 J/(mol*K), we have:
ΔT = (2/3) * (10 kJ) / (3 * R) * (1000 J/1 kJ) = (2/3) * (10,000 J) / (3 * 8.314 J/(mol*K))
Simplifying further, we get:
ΔT = (2/3) * (10,000 J) / (3 * 8.314 J/(mol*K)) ≈ 1.83 x 10^5 K
Therefore, the temperature of the gas is approximately 1.83 x 10^5 K.
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An impulse internal to the system will not change the momentum of
that system ( True or False)
False. An impulse internal to the system can change the momentum of that system.
According to Newton's third law of motion, every action has an equal and opposite reaction. When an impulse occurs within a system, it involves the application of an internal force for a certain period of time, resulting in a change in momentum. The impulse-momentum principle states that the change in momentum of an object is equal to the impulse applied to it. Therefore, an impulse internal to the system can indeed cause a change in the momentum of the system.
For example, in a collision between two objects, such as billiard balls on a pool table, the impulses exerted between the balls during the collision will cause their momenta to change. The change in momentum is a result of the internal forces between the objects during the collision. This demonstrates that an impulse internal to the system can alter the momentum of the system as a whole.
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An inclined plane forms an angle of inclination of 30 degrees with a horizontal plane. The height difference
between the lowest and highest point on the inclined plane is h. - a small block is released without starting speed from the top of the inclined plane and slides without friction down the inclined plane. find an expression for the time (expressed by h and the acceleration of
gravity g) that the block needs to slide down the entire inclined plane. - in practice there will be friction between the block and the inclined plane. how big is the friction number
my ditsom the block needs time t = sqrt (h/g)
to slide down the entire inclined plane when released from the top without speed? -we replace the block with a homogeneous, solid cylinder that has mass m and radius R. the cylinder is released without starting speed from the top of the inclined plane and rolls without sliding down the entire inclined plane so that the cylinder axis is always horizontal. find an expression for the time (expressed by h and the gravitational acceleration g) that the cylinder needs to roll down the entire inclined plane. Ignore
friction work.
The energy conservation approach used for the block does not directly apply to the rolling cylinder
To find the expression for the time it takes for the block to slide down the inclined plane without friction, we can use the concept of conservation of energy.
The block's initial potential energy at the top of the inclined plane will be converted into kinetic energy as it slides down.
Without friction:
The potential energy (PE) at the top of the inclined plane is given by:
[tex]PE = mgh[/tex]
where m is the mass of the block, g is the acceleration due to gravity, and h is the height difference between the lowest and highest point on the inclined plane.
The kinetic energy (KE) at the bottom of the inclined plane is given by:
[tex]KE = (1/2)mv^2[/tex]
where v is the final velocity of the block at the bottom.
According to the principle of conservation of energy, the potential energy at the top is equal to the kinetic energy at the bottom:
[tex]mgh = (1/2)mv^2[/tex]
We can cancel out the mass (m) from both sides of the equation, and rearrange to solve for the final velocity (v):
[tex]v = sqrt(2gh)[/tex]
The time (t) it takes for the block to slide down the entire inclined plane can be calculated using the equation of motion:
[tex]s = ut + (1/2)at^2[/tex]
where s is the height difference, u is the initial velocity (which is zero in this case), a is the acceleration (which is equal to g), and t is the time.
Since the block starts from rest, the initial velocity (u) is zero, and the equation simplifies to:
[tex]s = (1/2)at^2[/tex]
Substituting the values of s and a, we have:
[tex]h = (1/2)gt^2[/tex]
Solving for t, we get the expression for the time it takes for the block to slide down the entire inclined plane without friction:
[tex]t = sqrt(2h/g)[/tex]
With friction:
To determine the frictional force acting on the block, we need additional information about the block's mass, coefficient of friction, and other relevant factors.
Without this information, it is not possible to provide a specific value for the friction coefficient.
Solid Cylinder Rolling Down:
If a homogeneous solid cylinder is released from the top of the inclined plane and rolls without sliding, the analysis becomes more complex.
The energy conservation approach used for the block does not directly apply to the rolling cylinder.
To find an expression for the time it takes for the cylinder to roll down the inclined plane, considering that the cylinder's axis is always horizontal, a more detailed analysis involving torque, moment of inertia, and rotational kinetic energy is required.
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A cylindrical copper cable carries a current of 1200 A. There is a potential difference of 0.016 V between two points on the cable that are 0.24 m apart. What is the diameter the cable? The resistivity of copper is 1.7 x 10^-8 Ωm.
A cylindrical copper cable carries a current of 1200 A. There is a potential difference of 0.016 V between two points on the cable that are 0.24 m apart.
The resistivity of copper is 1.7 x 10^-8 Ωm.
The formula for resistance is:
R = (ρl)/AR is resistanceρ is resistivity l is the length of the wireA is cross-sectional area of wire, the formula for cross-sectional area is:
[tex]A = (ρl)/RA = (ρl)/R= (1.7 x 10^-8 Ωm * 0.24 m)/((0.016 V)/1200 A))A = 5.1 x 10^-6 m^2[/tex]
Now, using the formula for cross-sectional area of a cylinder:
[tex]A = πd²/4We can write: πd²/4 = 5.1 x 10^-6 m^2d² = (4 * 5.1 x 10^-6 m^2)/πd² = 1.63 x 10^-6 m²d = √(1.63 x 10^-6 m²)d = 1.28 x 10^-3 m = 1.28 mm,[/tex]
the diameter of the copper cable is 1.28 mm.
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Using Gauss' law, obtain in every universe (o Spsco): the profile of the electric field density vector D(p), determine electric flux v(), the resulting electric field vector E(p) for a charge distributed on a spherical shell of inner radius p=a
р and outer radius q=d. whose distribution is =
pvQI(41p (b-a)) [C/m3] at the origin of the coordinates. Draw the Gaussians correctly to obtain the solution for each part of the problem space. Draw the profile of the flux, and the electric field for all environments.
To solve this problem using Gauss' law, let's consider the charge distribution on the spherical shell between inner radius p=a and outer radius q=d. The charge density distribution is given by ρ = pvQI(4πp(b-a)) [C/m³] at the origin of the coordinates.
First, we'll determine the electric field density vector D(p) using Gauss' law. Gauss' law states that the electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of the medium.
Since we have a spherical symmetry in this problem, we'll consider a Gaussian surface in the form of a sphere with radius r. We'll calculate the electric flux through this Gaussian surface and equate it to the total charge enclosed.
The resulting electric field vector E(p) is related to D(p) by the equation E = εD, where ε is the permittivity of the medium.
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8. A 5.00−kg bowling ball moving at 8.00 m/s collides with a 0.850−kg bowling pin, which is scattered at an angle to the initial direction of the bowling ball and with a speed of 15.0 m/s. a. Calculate the final velocity (magnitude and direction) of the bowling ball. Answer b. Is the collision elastic? Answer 9. A wheel rotates at a constant rate of 2.0×10 3 rev/min. (a) What is its angular velocity in radians per second? Answer (b) Through what angle does it turn in 10 s? Express the solution in radians and degrees. Answer Radians Answer Degrees. 10. A wheel has a constant angular acceleration of 7.0rad/s 2 . Starting from rest, it turns through 400rad. (a) What is its final angular velocity? Answer (b) How much time elapses while it turns through the 400 radians? Answer
The angular velocity of the wheel is 209.44 radians/s.the final velocity of the bowling ball is 36.67 m/s in the positive direction.
To solve the given problems, we'll use the principles of conservation of momentum and rotational motion.8a. Calculate the final velocity (magnitude and direction) of the bowling ball:
Let's assume the positive direction is the initial direction of the bowling ball. According to the law of conservation of momentum:
(mass of bowling ball) × (initial velocity of bowling ball) = (mass of bowling pin) × (final velocity of bowling ball) + (mass of bowling pin) × (final velocity of bowling pin)(5.00 kg) × (8.00 m/s) = (0.850 kg) × (final velocity of bowling ball) + (0.850 kg) × (15.0 m/s) 40.00 kg·m/s = 0.7225 kg·m/s + 12.75 kg·m/s + (0.7225 kg) × (final velocity of bowling ball)
Simplifying the equation:
40.00 kg·m/s - 13.4725 kg·m/s = (0.7225 kg) × (final velocity of bowling ball) 26.5275 kg·m/s = (0.7225 kg) × (final velocity of bowling ball)
final velocity of bowling ball = 26.5275 kg·m/s / 0.7225 kg
final velocity of bowling ball = 36.67 m/s
Therefore, the final velocity of the bowling ball is 36.67 m/s in the positive direction.
8b. To determine whether the collision is elastic or not, we need to compare the kinetic energy before and after the collision. If the kinetic energy is conserved, the collision is elastic. If not, it is inelastic.
Kinetic energy before the collision:
KE_initial = (1/2) × (mass of bowling ball) × (initial velocity of bowling ball)^2
= (1/2) × (5.00 kg) × (8.00 m/s)^2
= 160 J
Kinetic energy after the collision:
KE_final = (1/2) × (mass of bowling ball) × (final velocity of bowling ball)^2 + (1/2) × (mass of bowling pin) × (final velocity of bowling pin)^2
= (1/2) × (5.00 kg) × (36.67 m/s)^2 + (1/2) × (0.850 kg) × (15.0 m/s)^2
= 3368 J
Since KE_initial = 160 J and KE_final = 3368 J, the kinetic energy is not conserved, indicating an inelastic collision.
9a. Given:
Angular velocity = 2.0 × 10^3 rev/min
To convert rev/min to radians per second, we need to use conversion factors:
1 revolution (rev) = 2π radians
1 minute (min) = 60 seconds (s)
Angular velocity = (2.0 × 10^3 rev/min) × (2π radians/1 rev) × (1 min/60 s)
= (2.0 × 10^3) × (2π/60) radians/s
= 209.44 radians/s
Therefore, the angular velocity of the wheel is 209.44 radians/s.
Given:
Time = 10 s
Using the formula for angular displacement:
θ = ω_initial × t + (1/2) × α × t^2
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7. The steady state and pseudosteady state flow equations in a circular drainage area, and productivity Index are given as q=7.081*10¯^3 kh/Bμ (rhoe-rhowf)/({In(re/rw)+s}) q=7.081*10¯^3 kh/Bμ (p-rhowf)/({In(re-rw)-0,75+s}) J=q/Δp Similary the dimensionless pressure, dimensionless rate and dimensionless productivity index are defined as: pn=Δp/pch qn=q/qch Jn=J/Jch
a. You are asked to find out the what are the characteristic variables that make those varaibles dimensionless and write the dimensionless pressure, rate and productivity index variables? b. Also find out how do these three dimensionless variables relate to each other? Or aren't they related at all?
a. The characteristic variables that make those variables dimensionless and write the dimensionless pressure, rate, and productivity index variables are as follows:Dimensionless Pressure (pn):
(Δp/pch)Dimensionless Rate (qn): (q/qch)Dimensionless Productivity Index (Jn): (J/Jch)The characteristic variables (pch, qch, and Jch) are obtained by choosing appropriate reference values for pressure (pch), rate (qch), and productivity index (Jch).b. These three dimensionless variables are related by the equationJn = pn/qnProductivity index (J) is related to pressure (p) and rate (q) through the following equation:
J = q/ΔpFor dimensionless variables, we divide both sides of the above equation by qch/Jch, which gives usJn = pn/qnThus, the dimensionless productivity index is equal to the dimensionless pressure divided by the dimensionless rate.About Characteristic variablesCharacteristic variables come from experimental observations or obtained from experimental intuition on the process.
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When you are looking at a rainbow the Sun is located: Right in front of you The location of the Sun could be anywhere Right behind you At a 90 degree angle relative to your location
when you look at a rainbow, the sun is located right behind you, at a 42-degree angle relative to your location. The sun's position is critical in creating the rainbow, and it is a fascinating meteorological phenomenon that never ceases to amaze us.
When you look at a rainbow, the sun is located at a 42-degree angle relative to your location. Rainbows are a meteorological phenomenon that occurs when sunlight enters water droplets and then refracts, reflects, and disperses within the droplets.
A primary rainbow is caused by a single reflection of sunlight within the water droplets, whereas a secondary rainbow is caused by two internal reflections of light within the droplets.
To locate the sun's position concerning a rainbow, consider the following. When you see a rainbow, the sunlight enters the water droplets from behind your back and then disperses into the spectrum of colors.
Therefore, the sun is always behind you when you face a rainbow, as the sun's rays are reflected off the raindrops and into your eyes.
However, the sun's angle relative to the observer is crucial in creating a rainbow.
The sun's position can be determined using the following formula:
The light enters the droplets at a 42-degree angle from the observer's shadow and then leaves the droplets at a 42-degree angle, creating the arc shape that you see.
In conclusion, when you look at a rainbow, the sun is located right behind you, at a 42-degree angle relative to your location.
The sun's position is critical in creating the rainbow, and it is a fascinating meteorological phenomenon that never ceases to amaze us.
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A 2nC charge is located at (0,−1)cm and another 2nC charge is located at (−3,0)cm. What would be the magnitude of the net electric field at the origin (0,0)cm ?
The magnitude of the net electric field at the origin (0,0)cm due to two point charges located at (0, -1)cm and (-3, 0)cm, each with a charge of 2nC, is 1.85 x 10⁸ N/C.
To determine the magnitude of the net electric field at the origin (0,0)cm due to two point charges located at (0, -1)cm and (-3, 0)cm, each with a charge of 2nC, we can make use of Coulomb's Law and vector addition.
The magnitude of the electric field at any point in space is given by:
E= kq/r²Where k is Coulomb's constant (9 x 10⁹ Nm²/C²), q is the charge, and r is the distance between the point charge and the point where the electric field is being measured. The electric field is a vector quantity and is directed away from a positive charge and towards a negative charge.
To determine the net electric field at the origin (0,0)cm due to the two charges, we can calculate the electric field due to each charge individually and then add them vectorially. We can represent the electric field due to the charge at (0,-1)cm as E1 and the electric field due to the charge at (-3,0)cm as E2.
The distance between each charge and the origin is given by: r1 = 1 cm r2 = 3 cm Now, we can calculate the magnitude of the electric field due to each charge:
E1 = (9 x 10⁹ Nm²/C²) * (2 x 10⁻⁹ C) / (1 cm)² = 1.8 x 10⁸ N/C
E2 = (9 x 10⁹ Nm²/C²) * (2 x 10⁻⁹ C) / (3 cm)² = 4 x 10⁷ N/C
Now, we need to add the two electric fields vectorially. To do this, we need to consider their directions. The electric field due to the charge at (0,-1)cm is directed along the positive y-axis, whereas the electric field due to the charge at (-3,0)cm is directed along the negative x-axis.
Therefore, we can represent E1 as (0, E1) and E2 as (-E2, 0).The net electric field is given by:E_net = √(Ex² + Ey²)where Ex and Ey are the x and y components of the net electric field.
In this case,Ex = -E2 = -4 x 10⁷ N/CEy = E1 = 1.8 x 10⁸ N/C
Hence,E_net = √((-4 x 10⁷)² + (1.8 x 10⁸)²) = 1.85 x 10⁸ N/CTo summarize, the magnitude of the net electric field at the origin (0,0)cm due to two point charges located at (0, -1)cm and (-3, 0)cm, each with a charge of 2nC, is 1.85 x 10⁸ N/C.
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A new communications satellite launches into space. The rocket carrying the satellite has a mass of 2.35 * 10^6 kg . The engines expel 3.55 * 10^3 kg of exhaust gas during the first second of liftoff giving the rocket an upwards velocity of 5.7 m/s.
At what velocity is the exhaust gas leaving the rocket engines?
Ignore the change in mass due to the fuel being consumed. The exhaust gas needed to counteract the force of gravity is accounted for, and should not be part of this calculation. Show all calculations.
The mass of the rocket is 2.35 x 10^6 kg. The mass of the exhaust gas expelled in 1 second is 3.55 x 10^3 kg.
The initial velocity of the rocket is 0 m/s. The final velocity of the rocket after 1 second of lift off is 5.7 m/s. At what velocity is the exhaust gas leaving the rocket engines? We can calculate the velocity at which the exhaust gas is leaving the rocket engines using the formula of the conservation of momentum.
The equation is given as:m1u1 + m2u2 = m1v1 + m2v2Where m1 and m2 are the masses of the rocket and exhaust gas, respectively;u1 and u2 are the initial velocities of the rocket and exhaust gas, respectively;v1 and v2 are the final velocities of the rocket and exhaust gas, respectively.
Multiplying the mass of the rocket by its initial velocity and adding it to the mass of the exhaust gas multiplied by its initial velocity, we have:m1u1 + m2u2 = 2.35 x 10^6 x 0 + 3.55 x 10^3 x u2 = m1v1 + m2v2Next, we calculate the final velocity of the rocket.
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a
cylinder of radius .35 m is released from rest to roll down a
frictionless slope, the cylinder has a velocity of 4.85 m/s. what
vertical height did the cylinder start from?
The principle of conservation of mechanical energy states that in a closed system where only conservative forces (such as gravity or elastic forces) are acting, the total mechanical energy remains constant over time. The cylinder started from a vertical height of approximately 0.621 meters.
To determine the vertical height from which the cylinder started, we can use the principle of conservation of mechanical energy. The mechanical energy of the cylinder is conserved as it rolls down the frictionless slope, so the initial potential energy is equal to the final kinetic energy.
The potential energy (PE) of the cylinder at the initial height can be calculated using the formula:
[tex]PE = m * g * h[/tex]
where m is the mass of the cylinder, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height.
The kinetic energy (KE) of the cylinder at the final velocity can be calculated using the formula:
[tex]KE = (1/2) * I * \omega^2[/tex]
where I is the moment of inertia of the cylinder and ω is the angular velocity.
For a solid cylinder rolling without slipping, the moment of inertia can be expressed as:
[tex]I = (1/2) * m * r^2[/tex]
where r is the radius of the cylinder.
Since the cylinder is released from rest, the initial velocity is 0 m/s, and thus the initial kinetic energy is also 0.
Setting the initial potential energy equal to the final kinetic energy, we have:
[tex]m * g * h = (1/2) * I * \omega^2[/tex]
Substituting the expressions for I and ω, we get:
[tex]m * g * h = (1/2) * (1/2) * m * r^2 * (v/r)^2[/tex]
Simplifying the equation, we have:
[tex]g * h = (1/4) * v^2[/tex]
Solving for h, we find:
[tex]h = (1/4) * v^2 / g[/tex]
Substituting the given values, we can calculate the vertical height:
[tex]h = (1/4) * (4.85 m/s)^2 / 9.8 m/s^2[/tex]
[tex]h = 0.621 m[/tex]
Therefore, the cylinder started from a vertical height of approximately 0.621 meters.
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Q|C (a) Find the number of moles in one cubic meter of an ideal gas at 20.0°C and atmospheric pressure.
The number of moles in one cubic meter of an ideal gas at 20.0°C and atmospheric pressure is approximately 44.62 moles.
To calculate the number of moles in a gas, we can use the ideal gas law equation,
PV = nRT
Where,
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature in Kelvin
At atmospheric pressure, the standard pressure is approximately 101.325 kPa or 101325 Pa. We convert this pressure to the SI unit of Pascal (Pa). Using the ideal gas law, we can rearrange the equation to solve for the number of moles (n),
n = PV / RT
The temperature is given as 20.0°C. We need to convert it to Kelvin by adding 273.15,
T = 20.0°C + 273.15 = 293.15 K
Now we have all the values needed to calculate the number of moles. The ideal gas constant, R, is approximately 8.314 J/(mol·K).
Plugging in the values,
n = (101325(1)/(8.314/293.15)
n ≈ 44.62 moles
Therefore, the number of moles in one cubic meter of an ideal gas at 20.0°C and atmospheric pressure is approximately 44.62 moles.
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"How much would a simple pendulum deflect due to the
gravity of a nearby mountain? As a model of a large mountain, use a
sphere of radius R = 2.5 km and mass density = 3000 kg/m³.
Due to the gravity of the sphere, the deflection of the simple pendulum will be greater.
A simple pendulum is a swinging object that oscillates back and forth around a stable equilibrium position. Its motion is used to explain gravity and to determine the gravitational force. The force of gravity on the Earth is a crucial factor for the simple pendulum's motion. The pendulum's deflection can be computed with the formula:
T = 2π * √(l/g) Where
T is the period of the pendulum
l is the length of the pendulum's support string
g is the acceleration due to gravity
Due to the gravity of a nearby mountain, a simple pendulum would deflect.The magnitude of the gravitational force at any point on the sphere's surface is given by:
F = (G * m * M) / R² Where
F is the gravitational force
G is the gravitational constant
m is the mass of an object
M is the mass of the sphere
R is the sphere's radius
Due to the gravitational force of the sphere, the deflection of the pendulum will be greater.
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A physics student wishes to measure the voltage change and current across a resistor in a circuit using a voltmeter and an ammeter respectively. How should the student connect the voltmeter and ammeter to the circuit? O a. The voltmeter should be connected in series with the resistor, and the ammeter should be connected in parallel with the resistor. O b. The voltmeter should be connected in series with the resistor, and the ammeter should be connected in series with the resistor. O c. The voltmeter and ammeter should be connected in a series combination that is, in turn, connected in parallel with the resistor d. The voltmeter should be connected in parallel with the resistor, and the ammeter should be connected in parallel with the resistor. Oe. The voltmeter should be connected in parallel with the resistor, and the ammeter should be connected in series with the resistor. QUESTION 17 A conducting, multi-turn circular loop of radius 12.0 cm carries a current of 15.0 A and has a magnetic field strength of 0.0250 T at the center of the loop. How many turns are in the loop? O a. 160 turns O b.583 turns O c. 274 turns O d. 515 turns O e. 318 turns QUESTION 18 3.0 moles of helium gas, that initially occupies a volume of 30 L at a temperature of 280 K, isothermally expands to 40 L. How much work does the gas perform on its environment? O a. 2.00 kcal O b.5.00 kcal O c. 6.00 kcal O d. 3.00 kcal O e. 4.00 kcal
Answer: While measuring voltage change and current across a resistor in a circuit, a physics student should connect the voltmeter in parallel to the resistor, and the ammeter in series with the resistor.
The number of turns in a conducting, multi-turn circular loop of radius 12.0 cm that carries a current of 15.0 A and has a magnetic field strength of 0.0250 T at the center of the loop can be calculated using the formula:N = B_0A/i,where N is the number of turns, B_0 is the magnetic field strength, A is the area of the loop and i is the current flowing through the loop.
Area of the circular loop, [tex]A = πr² = π(0.12 m)² = 0.045 m[/tex]
The moles of helium gas that initially occupies a volume of 30 L at a temperature of 280 K and isothermally expands to 40 L can be calculated using the ideal gas law formula, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.
Rearranging the formula to get the number of moles of gas:[tex]n = PV/RT[/tex]
The work done by the gas can be calculated using the formula, [tex]W = nRT ln(V_f/V_i), where V_f[/tex] is the final volume and V_i is the initial volume.
The work done is given by:[tex]W = 3.0 mol x (8.314 J/mol K) x 280 K ln(40/30)W = 2.01 kJ = 2.01/4.18 = 0.481 kcal[/tex]
Therefore, the work done by the gas on its environment is 0.481 kcal.
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Four 700 gram masses are the four corners of a square with sides of 50.0 centimeters. Find the gravitational force on one mass as a result of the other three. G = 6.67 * 10^-11 Nm^2/kg^2.
The gravitational force on one mass as a result of the other three is 3.27 x 10⁻¹⁰ N.
What is the gravitational mass on one mass?The gravitational force on one mass as a result of the other three is calculated by applying the following formula;
F = Gm₁m₄/r₁₄² + Gm₂m₄/r₂₄² + Gm₃m₄/r₃₄²
F = G[m₁m₄/r₁₄² + m₂m₄/r₂₄² + m₃m₄/r₃₄²]
where;
G is the universal gravitational constantr is the distance between the massThe distance between the masses are equal, except the two masses on the opposite diagonal.
the distance on opposite diagonal = r₁₄
r₁₄ = √(50² + 50²)
r₁₄ = 70.71 cm = 0.707 m
The gravitational force on one mass as a result of the other three is calculated as;
F = G[m₁m₄/r₁₄² + m₂m₄/r₂₄² + m₃m₄/r₃₄²]
m₁ = m₂ = m₃ = m₄ = 0.7 kg
F = Gm²(1/r₁₄² + 1/r₂₄² + 1/r₃₄²)
F = 6.67 x 10⁻¹¹ x (0.7²) [1/0.707² + 1/0.5² + 1/0.5²]
F = 3.27 x 10⁻¹⁰ N
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the back of the upper arm extends the forearmThis muscle in a professional boxer exerts a force of 1.46 * 10 ^ 3 N with an effective perpendicular lever arm of producing acceleration of the forearm of 121rid / (s ^ 2) What is the moment of of the boxer's forearm? Note the perpendicular lever is defined ) so that the magnitude of torque
The moment of a boxer's forearm is determined using the following formula:
τ = r × F × sin(θ)Where :r is the effective perpendicular lever arm,
F is the force exerted by the muscle in a professional boxerθ is the angle between the force vector and the direction of the lever armτ is the torque produced by the muscle in a professional boxer Given:
F = 1.46 × 10³ N, r = 121 m/s²sin(θ) = 1 (since the angle between r and F is 90°)
τ = 121 × 1.46 × 10³ × 1τ = 177,660 Nm
the moment of the boxer's forearm is 177,660 Nm.
The formula for torque or moment is τ = r × F × sin(θ)
where r is the effective perpendicular lever arm, F is the force exerted by the muscle in a professional boxer, θ is the angle between the force vector and the direction of the lever arm, τ is the torque produced by the muscle in a professional boxer.
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Question 3 (4 points) A nano-satellite has the shape of a disk of radius 0.70 m and mass 20.25 kg. The satellite has four navigation rockets equally spaced along its edge. Two navigation rockets on opposite sides of the disk fire in opposite directions to spin up the satellite from zero angular velocity to 10.5 radians/s in 22.5 seconds. If the rockets each exert their force tangent to the edge of the satellite (the angle theta between the force and the radial line is 90 degrees), what was is the force of EACH rocket, assuming they exert the same magnitude force on the satellite? Your Answer: Answer units
A nano-satellite has the shape of a disk of radius 0.70 m and mass 20.25 kg. The satellite has four navigation rockets equally spaced along its edge. the force exerted by EACH rocket is 0 N.
To find the force exerted by each rocket, we can use the principle of conservation of angular momentum.
The angular momentum of the satellite can be expressed as the product of its moment of inertia and angular velocity:
L = Iω
The moment of inertia of a disk can be calculated as:
I = (1/2) * m * r^2
Given:
Radius of the satellite (disk), r = 0.70 m
Mass of the satellite (disk), m = 20.25 kg
Angular velocity, ω = 10.5 rad/s
We can calculate the moment of inertia:
I = (1/2) * m * r^2
= (1/2) * 20.25 kg * (0.70 m)^2
Now, we can determine the initial angular momentum of the satellite, which is zero since it starts from rest:
L_initial = 0
The final angular momentum of the satellite is given by:
L_final = I * ω
Since the two rockets on opposite sides of the disk fire in opposite directions, the net angular momentum contributed by these rockets is zero. Therefore, the final angular momentum is only contributed by the other two rockets:
L_final = 2 * (Force * r) * t
where:
Force is the force exerted by each rocket
r is the radius of the satellite (disk)
t is the time taken to reach the final angular velocity
Setting the initial and final angular momenta equal, we have:
L_initial = L_final
0 = 2 * (Force * r) * t
Simplifying the equation, we can solve for the force:
Force = 0 / (2 * r * t)
= 0
Therefore, the force exerted by EACH rocket is 0 N.
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Given an object distance of 12 cm and a lens with focal length
of magnitude 4 cm, what is the image distance for a convex lens?
Give your answers in cm.
The answer is the image distance for a convex lens is 6 cm. Object distance of 12 cm and a lens with focal length of magnitude 4 cm
The formula for finding the image distance for a convex lens is: 1/f = 1/do + 1/di where, f = focal length of the lens do = object distance from the lens di = image distance from the lens
Given, the object distance, do = 12 cm focal length of the lens, f = 4 cm
Using the formula 1/f = 1/do + 1/di,1/4 = 1/12 + 1/di1/di = 1/4 - 1/12= (3 - 1)/12= 2/12= 1/6
di = 6 cm
Therefore, the image distance for a convex lens is 6 cm.
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The following three questions relate to the information here: Ripples radiate out from vibrating source in water. After 6.00 s, 42 ripples have been generated with the first ripple covering a distance of 3.00 m from the source (each ripple constitutes a wave).
What is the wavelength of the ripples? (a) 0.048 m (b) 0.071 m (c) 0.43 m (d) 3.0 m
What is the frequency of the ripples? (a) 14 Hz (b) 7.0 Hz (c) 0.33 Hz (d) 0.17 Hz
What is the speed of the ripples? (a) 0.1 m s−1 (b) 0.2 m s−1 (c) 0.4 m s−1 (d) 0.5 m s
The correct answers to the given questions are as follows:
a) The wavelength of the ripples is (d) 3.0 m.
b) The frequency of the ripples is (b) 7.0 Hz.
c) The speed of the ripples is not provided in the given options. It is 21.0 m/s.
To solve these questions, we can use the formula:
v = λf,
where
v is the speed of the ripples,
λ is the wavelength, and
f is the frequency.
Wavelength of the ripplesGiven that the first ripple covers a distance of 3.00 m from the source, we can assume this is equal to the wavelength of the ripples:
λ = 3.00 m.
Therefore, the answer is (d) 3.0 m.
Frequency of the ripplesWe are given that after 6.00 seconds, 42 ripples have been generated. The frequency (f) can be calculated by dividing the number of ripples by the time:
f = number of ripples/time.
f = 42 ripples / 6.00 s.
f = 7.0 Hz.
Therefore, the answer is (b) 7.0 Hz.
Speed of the ripplesUsing the formula v = λf, we can substitute the known values:
v = (3.00 m) × (7.0 Hz).
v = 21.0 m/s.
Therefore, the answer is none of the provided options. The speed of the ripples is 21.0 m/s.
Therefore,
a) The wavelength of the ripples is (d) 3.0 m.
b) The frequency of the ripples is (b) 7.0 Hz.
c) The speed of the ripples is not provided in the given options. It is 21.0 m/s.
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The wavelength of the ripples is 0.071 m. The answer is (b) 0.071 m. The frequency of the ripples is 7.0 Hz. The answer is (b) 7.0 Hz. The speed of the ripples is approximately 0.497 m/s. The answer is (d).
After 6.00 s, 42 ripples have been generated, with the first ripple covering a distance of 3.00 m from the source.
Each ripple constitutes a wave.
(a) To find the wavelength of the ripples:
Wavelength = Total Distance / Number of Ripples
Wavelength = 3.00 / 42
Wavelength = 0.071 m
Therefore, the wavelength of the ripples is 0.071 m. The answer is (b) 0.071 m.
(b) To find the frequency of the ripples:
Frequency = Number of Ripples / Total Time
Frequency = 42 / 6.00
Frequency = 7.0 Hz
Therefore, the frequency of the ripples is 7.0 Hz. The answer is (b) 7.0 Hz.
(c) To find the speed of the ripples:
Speed = 7.0 × 0.071
Speed = 0.497 m/s
Therefore, the speed of the ripples is approximately 0.497 m/s. The answer is (d).
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A uniform magnetic field B has a strength of 5.5 T and a direction of 25.0° with respect to the +x-axis. A proton (1.602e-19)is traveling through the field at an angle of -15° with respect to the +x-axis at a velocity of 1.00 ×107 m/s. What is the magnitude of the magnetic force on the proton?
The magnitude of the magnetic force on the proton is 4.31 × 10⁻¹¹ N.
Given values: B = 5.5 Tθ = 25°q = 1.602 × 10⁻¹⁹ VC = 1.00 × 10⁷ m/s Formula: The formula to calculate the magnetic force is given as;
F = qvBsinθ
Where ;F is the magnetic force on the particle q is the charge on the particle v is the velocity of the particle B is the magnetic field strengthθ is the angle between the velocity of the particle and the magnetic field strength Firstly, we need to determine the angle between the velocity vector and the magnetic field vector.
From the given data, The angle between velocity vector and x-axis;α = -15°The angle between magnetic field vector and x-axis;β = 25°The angle between the velocity vector and magnetic field vectorθ = 180° - β + αθ = 180° - 25° - 15°θ = 140° = 2.44346 rad Now, we can substitute all given values in the formula;
F = qvBsinθF
= (1.602 × 10⁻¹⁹ C) (1.00 × 10⁷ m/s) (5.5 T) sin (2.44346 rad)F
= 4.31 × 10⁻¹¹ N
Therefore, the magnitude of the magnetic force on the proton is 4.31 × 10⁻¹¹ N.
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