Answer:
Gravitational Potential Energy of both will be equal
Explanation:
The gravitational potential energy is given by the following formula:
[tex]P.E = mgh[/tex]
where,
P.E = Gravitational Potential Energy
m = mass of object
g = acceleration due to gravity
h = height of object
Therefore, the gravitational potential energy of an object depends only upon the height of the object and its mass.
Since, both masses and the heights achieved by both people is equal. Therefore, their gravitational potential energies will also be equal
What is the frequency of a wave with a wavelength of 36 m and a speed of 12 m/s?
3Hz
O .33HZ
O 432HZ
O 312HZ
Answer:
312hz
Explanation:
thanks me later
classify the simple machine knife
Answer:
Wedge
Explanation:
A knife is used to cut things into pieces or even cut two items apart.
Now, it is classified as a Wedge because a Wedge is defined as a simple machine that is used in pushing two objects apart. Some examples include axes, knives & chisels.
16x^2y^2-25a^2b^2
factorize the expression
Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16[tex]x^{2}[/tex][tex]y^{2}[/tex]-25[tex]a^{2}[/tex][tex]b^{2}[/tex]
4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field. What are the magnitude and direction of the electric field at the point in question
Answer:
[tex]E = -6 \ N/C[/tex]
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
Explanation:
From the question we are told that
The charge on the small object is [tex]Q = -4.00 \ nC = -4.00 *10^{-9} \ C[/tex]
The force is [tex]F = 24 \ nN = 24 *10^{-9} \ N[/tex]
Generally the magnitude of the electric field is mathematically represented as
[tex]E = \frac{F}{Q}[/tex]
=> [tex]E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}[/tex]
=> [tex]E = -6 \ N/C[/tex]
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]
A circuit with a 12 V battery and lamp has a current of 3 A. What is the resistance of the lamp?
A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the volume through the hole. Find the time required for the number of particles to decrease to N/2. Express your answer in terms of A, V, and v.
Answer:
[tex]\frac{V}{2av}[/tex]
Explanation:
From the question we are told that
Volume V
Contains N particles
Leaks from a small hole of area A
Generally the equation for Flow rate is given as
Volume Flow Rate [tex]V_r = A * v[/tex]
Mathematically we find the time taken to flow half way which is given by
[tex]\frac{(V/2)}{A*v}[/tex]
Therefore the time taken is
[tex]\frac{V}{2av}[/tex]
A raindrop charged to 10 μC experiences electric field of 10,000 V/m from a tip of a tree branch. What is the electric force acting on the raindrop? (1μC= 10^-6C)
a. 0.001 N
b. 100 N
c. 0.1N
d. 0.9 N
Answer:
(C). The magnitude of electric force acting on the raindrop is 0.1 N.
Explanation:
Given;
charge of the raindrop, Q = 10 μC = 10 x 10⁻⁶ C
electric field strength, E = 10,000 V/m
The electric force acting on the raindrop is given as;
F = EQ
where;
F is the electric force
Substitute the given values and solve for F.
F = (10,000)(10 x 10⁻⁶)
F = 0.1 N
Therefore, the magnitude of electric force acting on the raindrop is 0.1 N.
The y-component of the force F which a person exerts on the handle of the box wrench is known to be 70 lb. Determine the x-component and the magnitude of F.
Answer:
Fx = 121.24lb
F = 140lb
Explanation:
Since we are not given the angles subtended by the force, we can assume it to be 30 degrees.
The y component of the force expressed by the formula:
Fy = Fsintheta
Given the y-component of the force F to bee 70lb
70lb = Fsintheta
Get magnitude of the force
F = 70/sin theta
F = 70/sin 30
F = 70/0.5
F = 140lb
Get the x-component of the force
Fx = Fcos theta
Fx = 140cos 30
Fx = 140(0.8660)
Fx = 1,212.4lb
Hence the x-component of the force sis 121.24lb
Note that the angle used was assumed. Other values can as well be used
A car travels 60 km in the first 2 hours and 68 km in the next 2 hours. what is the cars average speed ?
Why is it easier to swim in sea than in a pond or river?
Answer:
Since sea water contains alot of salt the density of the sea water is higher as to the river water . So it makes it easier to swim in sea water
A car accelerates uniformly in a straight line with acceleration 10m/s 2 and travels 150m in a time interval of 5s. How far will it travel in the next 5s?
Explanation:
Given:
Acceleration of car = 10 m/s
Distance travelled in 5 sec = 150 m
To find:
Distance travelled in the next 5 seconds
Concept:
There are 2 ways to app this kind of questions .
Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.
Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.
Calculation:
v² = u² + 2as
=> ( u + at)² = u² + 2as
=> u² + 2uat + a²t² = u² + 2as
=> 2uat + (at)² = 2as
=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)
=> 100u = 3000 - 2500
=> 100u = 500
=> u = 5 m/s
Distance travelled in 10 seconds :
s = ut + ½at²
=> s = (5 × 10) + ½(10)(10)²
=> s = 50 + 500
=> s = 550 m
Distance travelled in the 2nd half will be :
d = 550 - 150
=> d = 400 m
So final answer is :
Why do we perform stork stand test
Answer:
umm becuase it is a test and you need them
Explanation:
why water in earthern pot remain cool in summer
Answer:
The water kept in an earthen pot seeps into the small pores in the pot and evaporates from the surface of the pot. The heat required for evaporation is taken from water inside the pot, thus cooling the water stored inside. This is the reason why on hot summer days water remains cool in earthen pot.
PLEASE MARK ME AS BRAINLIESTHow have telescopes changed the way scientists study our Solar System?
Answer:
Telescopes have changed the way scientists study the Solar System. Before telescopes, people were only able to observe the Universe with their naked eyes. ... If it were not for telescopes, scientists would not be able to detect possible threats to Earth or predict comet's sightings.
Explanation:
400 years ago, before telescopes, our understanding of the universe was very different.
This is what was believed:
We live on a spherical ball orbited by the rest of a finite, spherical universe.
Earth does not move. It is the center of the universe.
Our Sun orbits the Earth, as do all the other planets and the Moon.
The stars are distant objects, always perfect and unchanging.
How did telescopes & associated technologies unlock the secrets of the universe and
help us toward the understanding we have today where Earth is no longer at the center of
the universe? Instead, we know that ours is a small planet orbiting a star in the suburbs
of a large galaxy filled with billions of other stars and planets, surrounded by billions of
other galaxies becoming increasingly ever distant from each other by the expansion of
space.
This is the story of how telescopes continuously changed our understanding of the
universe and our place in it - transforming our view of our universe. And we still have
much more to discover!
2. Before telescopes, we could only use our eyes and a variety of measuring instruments to
plot the positions and movements of objects in the sky to create a limited understanding
of our universe. We had no way to know what these objects actually were and little
evidence for our relationship to the cosmos.
How much work, in kilojoules, went into thermal energy produced by friction?
The question is incomplete. Here is the complete question.
Suppose the mass of a loaded elevator is 1600 kg.
(a) What force, in Newtons, must be supplied by the elveator's cable to produce an acceleration of 0.745 m/s² upwards against a 185N frictional force?
(b) How much work, in joules, is done by the cable in lifting the elevator 21m?
(c) What is the final speed, in meters per second, of the elevator if it starts from rest?
(d) How much work, in kilojoules, went into thermal energy produced by friction?
Answer: (a) F = 1377 N
(b) W = 28917 J
(c) v = 5.6 m/s
(d) W = 3.885 kJ
Explanation:
(a) According to Newton's Second Law: [tex]F_{net}=m.a[/tex], in which, [tex]F_{net}[/tex] is the vetorial sum of all the forces in a system and its unit is [F] = kg.m/s² or newton (N).
In the elevator's case, and assuming going upwards is positive:
[tex]F-F_{f}=m.a[/tex]
F - 185 = 1600(0.745)
F = 1377 N
For an elevator to produce an acceleration of 0.745m/s² upwards, the cables have to supply a force of 1377 newtons.
(b) Work is the energy transferred to an object while is being moved. It is calculated as: W = F.s. Its unit is [W] = N.m or Joule (J)
In the elevator's cable:
W = 1377.21
W = 28917 J
The work done by the elevator's cable is W = 28917 joules.
(c) Acceleration is variation in velocity along time. Since we know the displacement of the elevator:
[tex]v^{2}=v_{0}^{2}+2a\Delta x[/tex]
where:
v₀ is initial velocity, which is this case v₀=0 because it starts from rest;
a is acceleration;
Δx is the displacement
Replacing values:
[tex]v^{2}=2(0.745)(21)[/tex]
[tex]v=\sqrt{31.29}[/tex]
v = 5.6 m/s
Final speed of the elevator is 5.6 m/s.
(d) [tex]W=F_{f}.s[/tex]
W = 185(21)
W = 3885 J
Work transferred into thermal energy because of friction is W = 3.885 kJ.
A tennis player on serve tosses a ball straight up. While the ball is in free fall, its acceleration does which of the following?
a. decreasesb. remains constant c. increases then decreasesd. increasese. decreases then increases
Answer:
acceleration remains constant at g= 9.8m/s²
velocity however ...increases
A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?
Answer:
h = 5.09 m
Explanation:
Applying the Law of conservation of energy to this situation, we can write:
[tex]Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}[/tex]
where,
h = height of the hill = ?
v = speed of cart at the end = 10 m/s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\[/tex]
h = 5.09 m
Find the vector representing the area of the rectangle with vertices and oriented so that it faces downward. The magnitude of the vector equals the magnitude of the area; the direction is perpendicular to the surface. Since there are two perpendicular directions, we pick one by giving an orientation for the surface
Answer:
The answer is "-72 k".
Explanation:
Please find the complete question in the attached file.
Given point:
[tex]A=(0,0,0)\\B=(0,8,0)\\C=(9,8,0)\\D=(9,0,0)[/tex]
[tex]\bar{AB} = (0i+8j+0k)-(0i+0j+0k)= 8j\\\\\bar{AC} = (9i+8j+0k)-(0i+0j+0k)= 9i+8j\\\\[/tex]
Calculating the area:
[tex]Area=\left|\begin{array}{ccc}i&j&k\\0&8&0\\9&8&0\end{array}\right|[/tex]
[tex]=i[8(0)-8(0)]-j[(0-0)]+k[(0-9(8))]\\\\=i[0-0]-j[(0)]+k[(0-72)]\\\\=i[0]-j[(0)]+k[(-72)]\\\\=-72 \ k[/tex]
A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?
Answer:
The work done by friction was [tex]-4.5\times10^{5}\ J[/tex]
Explanation:
Given that,
Mass of car = 1000 kg
Initial speed of car =108 km/h =30 m/s
When the car is stop by brakes.
Then, final speed of car will be zero.
We need to calculate the work done by friction
Using formula of work done
[tex]W=\Delta KE[/tex]
[tex]W=K.E_{f}-K.E_{i}[/tex]
[tex]W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2[/tex]
Put the value of m and v
[tex]W=0-\dfrac{1}{2}\times1000\times(30)^2[/tex]
[tex]W=-450000 \ J[/tex]
[tex]W=-4.5\times10^{5}\ J[/tex]
Hence, The work done by friction was [tex]-4.5\times10^{5}\ J[/tex]
The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity between t1 = 1.0s and t2 = 4.0s.
Answer: 15m/s
Explanation: Average Velocity is vector describing the total displacement of an object and the time taken to change its position. It is represented as:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
At t₁ = 1.0s, displacement x₁ is:
[tex]x(1)=25+3(1)^{2}[/tex]
x(1) = 28
At t₂ = 4.0s:
[tex]x(4)=25+3(4)^{2}[/tex]
x(4) = 73
Then, average speed is
[tex]v=\frac{73-28}{4-1}[/tex]
v = 15
The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s
An athlete swings a 5.00-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.800 m at an angular speed of 0.500 rev/s. What are its centripetal acceleration?
Answer:
0.32m/s2
Explanation:
obtained from a=wr2 where w=anular speed
r-radius
Centripetal acceleration refers to the acceleration experienced by an object moving in a circular path. The centripetal acceleration of the ball is approximately [tex]9.86 m/s^2.[/tex]
Centripetal acceleration is a result of the inward force known as the centripetal force that keeps the object moving in a curved path. This force is necessary to counteract the tendency of the object to move in a straight line tangent to the circle.
To find the centripetal acceleration of the ball, we can use the formula:
centripetal acceleration = [tex](angular speed)^2 * radius[/tex]
First, let's convert the angular speed from revolutions per second to radians per second. Since 1 revolution is equal to [tex]2\pi[/tex] radians, we have:
angular speed in radians per second = [tex]0.500 rev/s * 2\pi rad/rev[/tex]
angular speed in radians per second =[tex]\pi rad/s[/tex]
Now, we can substitute the values into the formula to calculate the centripetal acceleration:
centripetal acceleration = [tex](\pi rad/s)^2 * 0.800 m[/tex]
centripetal acceleration = [tex]\pi ^2 * 0.800 m[/tex]
Using the approximation[tex]\pi = 3.14[/tex], we can calculate the centripetal acceleration:
centripetal acceleration = [tex](3.14)^2 * 0.800 m[/tex]
centripetal acceleration =[tex]9.86 m/s^2[/tex]
Therefore, the centripetal acceleration of the ball is approximately [tex]9.86 m/s^2.[/tex]
For more details regarding centripetal acceleration, visit:
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A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is the magnitude of the current in the other wire
Answer:
The current is [tex]I_b = 400 \ A[/tex]
Explanation:
From the question we are told that
The length of the segment is [tex]l = 2.50 \ m[/tex]
The current is [tex]I_a = 1000 \ A[/tex]
The force felt is [tex]F = 4.0 \ N[/tex]
The distance of the second wire is [tex]d = 5.0 \ cm = 0.05 \ m[/tex]
Generally the current on the second wire is mathematically represented as
[tex]I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4 \pi * 10^{-7} \ N/A^2[/tex]
=> [tex]I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }[/tex]
=> [tex]I_b = 400 \ A[/tex]
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 39.0 cm/scm/s . What are:
a. The amplitude of the subsequent oscillations?
b. The block's speed at the point where x= 0.750 A?
Answer:
a
[tex]A = 0.081 \ m[/tex]
b
The value is [tex]u = 0.2569 \ m/s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.750 \ kg[/tex]
The spring constant is [tex]k = 17.5 \ N/m[/tex]
The instantaneous speed is [tex]v = 39.0 \ cm/s= 0.39 \ m/s[/tex]
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2[/tex]
Here a is the amplitude of the subsequent oscillations
=> [tex]A = \sqrt{\frac{m * v^ 2 }{ k} }[/tex]
=> [tex]A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }[/tex]
=> [tex]A = 0.081 \ m[/tex]
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2[/tex]
=> [tex]\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2[/tex]
=> [tex]u = 0.2569 \ m/s[/tex]
The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from rest at position x = 0 m and acquires a speed of 3.0 m / s after traveling the distance of 0.090 m shown above. What is the mass of the object?
Answer:
0.06 Kg
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Net Force (F) = 3 N
Mass (m) =?
Next, we shall determine the acceleration of the object. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Acceleration (a) =?
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Divide both side by 0.18
a = 9 / 0.18
a = 50 m/s²
Finally, we shall determine the mass of the object. This can be obtained as follow:
Net Force (F) = 3 N
Acceleration (a) = 50 N
Mass (m) =?
F = ma
3 = m × 50
Divide both side by 50
m = 3 / 50
m = 0.06 Kg
Therefore, the mass of the object is 0.06 Kg
The required value for the mass of object is 0.06 kg.
Given data:
The speed of object is, v = 3.0 m/s.
The distance travelled by object is, s = 0.090 m.
From the given graph in the question, we obtain the following details as,
The net force applied on the object is, [tex]F_{net}=3\;\rm N[/tex].
The initial velocity is, u = 0 m/s.
Now using the third kinematic equation of motion to obtain the acceleration of object as,
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Solving as,
a = 9 / 0.18
a = 50 m/s²
Now, as per the Newton's second law of motion, we have the expression for the applied force as,
[tex]F_{net}= ma[/tex]
Here, m is the mass of object.
Solving as,
[tex]3= m \times 50\\\\m = 0.06 \;\rm kg[/tex]
Thus, we can conclude that the required value for the mass of object is 0.06 kg.
Learn more about the Newton's second law here:
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A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same time, kinetic friction acts on the box as it slides so that it eventually stops when the pulling force stops. What is the magnitude of the frictional force
Answer:
The magnitude of the frictional force is 50 newtons.
Explanation:
The frictional force can be found as follows:
[tex] \Sigma F = ma [/tex]
[tex] F - F_{\mu} = ma [/tex]
Where:
F: is the applied force = 50 N
[tex]F_{\mu}[/tex]: is the frictional force
m: is the box's mass
a: is the acceleration
Since the box is pulled at a constant velocity, a = 0, so:
[tex] F - F_{\mu} = 0 [/tex]
[tex] F = F_{\mu} = 50 N [/tex]
Therefore, the magnitude of the frictional force is equal to the applied force, that is to say, 50 newtons.
I hope it helps you!
The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what is its velocity after 2 seconds?
Given that,
The acceleration of gravity is -9.8 m/s²
Initial velocity, u = 39.2 m/s
Time, t = 2 s
To find,
The final velocity of the shot.
Solution,
Let v is the final velocity of sling shot. Using first equation of motion to find it.
v = u +at
Here, a = -g
v = u-gt
v = (39.2)-(9.8)(2)
v = 19.6 m/s
So, its velocity after 2 seconds is 19.6 m/s.
A Light spiral spring is loaded with
a mass of 50g and it extends by
10cm.
(is calculate the period of
small vertical oscillations.
Answer:
0.63 s
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 g
Extention (e) = 10 cm
Period (T) =?
Next, we obtained 50 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
50 g = 50 g × 1 Kg / 1000 g
50 g = 0.05 kg
Next, we shall convert 10 cm to m. This is illustrated below:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass = 0.05 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 0.05 × 9.8
F = 0.49 N
Next, we shall determine the spring constant of the spring.
Extention (e) = 0.1 m
Force (F) = 0.49 N
Spring constant (K) =?
F = Ke
0.49 = K × 0.1
Divide both side by 0.1
K = 0.49 /0.1
K = 4.9 N/m
Finally, we shall determine the period as follow:
Mass = 0.05 Kg
Spring constant (K) = 4.9 N/m
Pi (π) = 3.14
Period (T) =?
T = 2π√(m/k)
T = 2 × 3.14 × √(0.05 / 4.9)
T = 6.28 × √(0.05 / 4.9)
T = 0.63 s
Thus, the period of oscillation is 0.63 s
Q1: An object is in a uniform circular motion. Which of the following statements must be
true?
Answer:
it has constant angular velocity
Explanation:
an object moving in a uniform circular motion moves with constant angular velocity about a point
A bag of shells weighs 1.5 N. What is its mass? approximately 1.5 kilograms approximately 1.5 pounds approximately 150 grams approximately 1.5 grams
Answer:
approximately 150 grams
Explanation:
The weight of an object is a product of its mass and acceleration of gravity acting upon it. Since weight (W) is force, it is measured in Newtons (N) or Kgms-²
W = mg
Where;
W = weight (N)
m = mass (kg)
g = acceleration due to gravity
g is a constant, which is approximately 10m/s²
Therefore, according to this question, the mass of a bag of shells that weighs 1.5N can be calculated thus:
m = W/g
m = 1.5/10
m = 0.15 kg
Converting 0.15kg to grams, we multiply by 1000 i.e 0.15 × 1000 = 150 grams.
Hence, the mass of the object is approximately 150grams.