Determine the cross-sectional area of an aluminum wire if its resistance is 0.1 ohm and its mass is 54 grams.

Answers

Answer 1
To determine the cross-sectional area of the aluminum wire, we need to use the following formula:

Resistance (R) = Resistivity (ρ) * Length (L) / Cross-sectional Area (A)

where ρ is the resistivity of the material (aluminum, in this case), L is the length of the wire, and A is its cross-sectional area.

We do not have the length of the wire, but we can use the mass of the wire to find it. The density of aluminum is 2.7 g/cm^3, so the volume of the wire is:

Volume = mass / density = 54 g / 2.7 g/cm^3 = 20 cm^3

Let's assume that the wire is a cylinder with length L and cross-sectional area A. Then the volume can also be expressed as:

Volume = L * A

Therefore, we can solve for the length of the wire:

L = Volume / A = 20 cm^3 / A

Substituting this into the resistance formula, we get:

0.1 ohm = (2.65 x 10^-8 ohm*m) * (20 cm^3 / A)

where the resistivity of aluminum is 2.65 x 10^-8 ohm*m.

Solving for the cross-sectional area, we get:

A = (2.65 x 10^-8 ohm*m) * (20 cm^3) / 0.1 ohm

A = 5.3 x 10^-6 m^2

Therefore, the cross-sectional area of the aluminum wire is approximately 5.3 x 10^-6 square meters.

Related Questions

suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. if the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is

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When 20 pillars of the same height are erected along the boundary of a circular stadium and the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, the total number of beams is 190.

There are 20 pillars, and each pillar can be connected to 16 non-adjacent pillars (i.e., every 2nd pillar). Hence, the total number of connections would be 20 × 16 = 320. However, each connection is counted twice since it connects two pillars. Therefore, we need to divide 320 by 2 to obtain the actual number of connections. Therefore, there are 160 beams that connect the top of the pillars.

However, we need to consider the beams that form the boundary of the stadium as well. Since there are 20 pillars, there are 20 beams that connect the adjacent pillars on the boundary. Therefore, the total number of beams would be 160 + 20 = 180. But we still need to consider the beams that connect the pillars at the center of the stadium, which would be the diameter of the circle on which the pillars are erected.

If the radius of the circle is r and the height of the pillars is h, then the length of the diameter would be 2r + h. Since there are 20 pillars, there would be 20 diameters. Hence, the total number of beams that connect the pillars at the center of the stadium would be 20 × (2r + h). Therefore, the total number of beams would be:180 + 20(2r + h)= 180+40r+20h. Thus, the total number of beams is 180 + 40r + 20h.

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Need answers asap!!! Pls and thank you!

Answers

Answer:

1) 0,11 A

2) 224 V

3) I added a photo of my drawing

4) 8,34 Ω

5) 0,94 A

6) 0,35 A

7) 0,94 A

Explanation:

1) Given:

R = 80 Ω

V = 9V

Find: I - ?

[tex]i = \frac{v}{r} = \frac{9}{80} ≈ 0.11[/tex]

.

2) Given:

I = 2 A

R1 = 100 Ω

R2 = 12 Ω

Find: V - ?

This circuit is connected in series

R = R1 + R2

R = 100 + 12 = 112 Ω

V = I × R

V = 2 × 112 = 224 V

.

3) I added a photo of my drawing

4) Given:

R1 = R2 = R3 = 82 Ω

R4 = 12 Ω

V = 9 V

Find: R (total) - ?

This circuit is connected in parallel

[tex]\frac{1}{r} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + \frac{1}{r4} [/tex]

Since R1 = R2 = R3, we can find the total resistance in these 3 resistors using this formula:

[tex]r = \frac{r1}{n} [/tex]

n - the number of resistors

r1 - the resistance of one resistor (when all resistors have the same resistance)

[tex]r(123) = \frac{82}{3} ≈ 27.33[/tex]

Now, let's find the remaining resistance:

[tex] \frac{1}{r(total)} = \frac{1}{27.33} + \frac{1}{12} = \frac{437}{3644} [/tex]

[tex]r(total) ≈8.34[/tex]

.

5) Given:

R1 = R2 = 50 Ω

R3 = 75 Ω

R4 = 45 Ω

V = 120V

Find: I - ?

R3 and R4 are connected in parallel

[tex] \frac{1}{r(34)} = \frac{1}{75} + \frac{1}{45} = \frac{8}{225} [/tex]

[tex]r(34) ≈28.13[/tex]

R1, R2 and R34 and conected in series

[tex]r(total) = 28.13 + 50 + 50 = 128.13[/tex]

[tex]i = \frac{v}{r} = \frac{120}{128.13} ≈0.94[/tex]

.

6) Given:

R3 = 75 Ω

Find: I3 - ?

First, we have to find the voltage across R3 and R4

In order to do that, first we have know what voltage is across R1 and R2 (since the resistance is these 2 resistors are the same, the voltage will also be the same):

V1 = V2 = I × R1 = 0,94 × 50 = 47 V

Then V4 = V5 (parallel conection) = 120 - 47 - 47 = 26 V

I3 = V3/R3 = 26/75 = 0,35 A

.

7) Given:

I (total) = 0,94 A

Find: I1, I2 - ?

I1 = I2 = 0,94 A, because the current in series connection stays the same in every part of the circuit

in a single slit experiment, what effect on the first two minima in the diffraction pattern would result as the slit width is decreased?

Answers

In a single-slit experiment, when the slit width is decreased, the first two minima in the diffraction pattern will spread out and move further away from the central maximum.

In a single-slit experiment, when light passes through a narrow slit, it diffracts and creates a pattern of bright and dark fringes on a screen placed behind the slit. The location of the fringes depends on the wavelength of the light, the distance between the slit and the screen, and the width of the slit.


1. The single-slit experiment is based on the principle of diffraction, where light passing through a narrow slit creates an interference pattern.
2. The diffraction pattern consists of a central maximum surrounded by alternating bright and dark fringes, called maxima and minima, respectively.
3. The positions of the minima are determined by the relationship: a * sin(θ) = m * λ, where 'a' is the slit width, 'θ' is the angle between the central maximum and the minima, 'm' is the order of the minima (1, 2, 3, etc.), and 'λ' is the wavelength of the light.
4. As the slit width 'a' decreases, the value of sin(θ) must increase to maintain equality, which means the angle 'θ' increases as well.
5. With an increased angle 'θ', the first two minima in the diffraction pattern move further away from the central maximum, causing the pattern to spread out.

Therefore, as the slit width is decreased in a single-slit experiment, the distance between the first two minima in the diffraction pattern will increase.

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A resultant force of 25 newtons act on a mass of 0.50kg starting from rest. find acceleration in meter per second squared

Answers

Explanation:

F = m* a

25 N = .50 kg * a

25/.50 = a = 50 m/s^2

a proton of kinetic energy 1.0 107 ev moves in a circular orbit in the magnetic field near the earth. the strength of the field is 5.0 105 t. what is the radius of the orbit?

Answers

The radius of the orbit is approximately 5.153 x [tex]10^{-2}[/tex] m or 5.15 cm.

To find the radius of the circular orbit of a proton with kinetic energy 1.0 x [tex]10^7[/tex] eV moving in a magnetic field of strength 5.0 x [tex]10^{-5}[/tex] T, follow these steps:
Convert the kinetic energy from electron volts (eV) to Joules (J) using the conversion factor

1 eV = 1.602 x [tex]10^{-19}[/tex] J.
Kinetic energy (J) = 1.0 x [tex]10^{7}[/tex] eV x 1.602 x [tex]10^{-19}[/tex] J/eV = 1.602 x [tex]10^{-12}[/tex] J
Calculate the velocity of the proton using the kinetic energy formula:
Kinetic energy = (1/2) x mass x [tex]velocity^2[/tex]
Solve for velocity:
Velocity = [tex]\sqrt{(2 x Kinetic energy / mass)}[/tex]
The mass of a proton is 1.67 x [tex]10^{-27}[/tex] kg

so:
Velocity = [tex]\sqrt{(2 \times 1.602 \times 10^{-12} J / 1.67 \times 10^{-27} kg)}[/tex]

Velocity  = 2.478 x [tex]10^{7}[/tex] m/s
Determine the charge of a proton, which is 1.602 x [tex]10^{-19}[/tex] C.
Calculate the radius of the orbit using the formula:
Radius = (mass x velocity) / (charge x magnetic field strength)
Radius = (1.67 x [tex]10^{-27}[/tex] kg x 2.478 x [tex]10^{7}[/tex] m/s) / (1.602 x[tex]10^{-19}[/tex]C x 5.0 x [tex]10^{-5}[/tex] T) = 5.153 x [tex]10^{-2}[/tex] m
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question 5 a certain non-conducing material has index of refraction equal to 1.65. what is the brewster's angle?

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As per the given information, a certain non-conducing material has an index of refraction equal to 1.65. We have to determine the Brewster's angle.

The Brewster's angle is defined as the angle of incidence at which the reflected light becomes completely polarized perpendicular to the plane of incidence. It is also known as the polarization angle.

According to the Snell's law of refraction, μ = sin i/sin rWhere, μ is the refractive index of the material, i is the angle of incidence, and r is the angle of refraction.The Brewster's angle is given by the formula: tan β = μWhere, β is the Brewster's angleNow, we have the value of the refractive index, which is equal to 1.65.

Hence, we can use this value to calculate the Brewster's angle as follows: tan β = 1.65β = tan⁻¹(1.65)β = 58.45°Therefore, the Brewster's angle is 58.45° for the given non-conducting material with a refractive index of 1.65.

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Please help !

A mercury thermometer is constructed as shown. The capillary tube has a diameter of 0.005 cm, and the bulb has a diameter of 0.31 cm. Neglecting the expansion of the glass, find the change in height of the mercury column for a temperature change of 31◦C. The volume expansion coefficient for mercury is 0.000182 (◦C)−1 .

Answer in units of cm.

Answers

With a temperature change of 31°C, the mercury column's height changes by 0.0106 cm.

How can you figure out the mercury rise in a thermometer?

As the temperature rises, the mercury will expand in the capillary tube, increasing its volume. V = VT = (1.8*10-4 (oC)-1)(0.100 cm3)(20 oC) = 3.6*10-4 cm3 = 0.36 mm3, or the formula V = VT.

[tex]A1 * h1 = A2 * h2[/tex]

[tex]A1 = πr1^2 = π(0.005 cm/2)^2 = 7.85 × 10^-5 cm^2[/tex]

The cross-sectional area of the bulb is:

[tex]A2 = πr2^2 = π(0.31 cm/2)^2 = 0.0755 cm^2[/tex]

[tex]ΔV = V0 * β * ΔT[/tex]

[tex]V0 = A1 * h1[/tex]

[tex]h2 = (A1/A2) * h1 + (ΔV/A2)[/tex]

[tex]h2 = (7.85 × 10^-5 cm^2)/(0.0755 cm^2) * h1 + (0.000182 (°C)^-1 * 31°C *[/tex] 7.85 × [tex]10^-5 cm^2)/(0.0755 cm^2)[/tex]

[tex]h2 = 0.0106 h1 + 0.00000122 cm[/tex]

[tex]Δh = h2 - h1 = 0.0106 h1 + 0.00000122 cm - h1 = 0.0106 cm[/tex]

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A circular hoop of mass m, radius r, and infinitesimal thickness rolls without slipping down a ramp inclined at an angle θ with the horizontal. a. what is the acceleration αα of the center of the hoop? b. what is the minimum coefficient of (static) friction μminμmin needed for the hoop to roll without slipping?

Answers

The acceleration (α) of the center of the hoop: α = (mg * sin(θ)) / (2m) = (g * sin(θ)) / 2

The minimum coefficient of static friction (μmin) needed for the hoop to roll without slipping: μmin = α / (g * cos(θ)) = (g * sin(θ) / 2) / (g * cos(θ)) = tan(θ) / 2

The acceleration (α) of the center of the circular hoop can be determined by applying Newton's second law and the principle of conservation of angular momentum. Considering gravitational force (mg), normal force (N), and friction force (f) acting on the hoop, we can write:

mg * sin(θ) - f = m * α (1) (linear motion) and f * r = I * α/r (2) (angular motion)

For a hoop, the moment of inertia (I) is given by I = m * r^2. Substituting this into equation (2) and solving for f, we get:

f = m * α (3)

Now, substitute equation (3) into equation (1):

mg * sin(θ) - m * α = m * α

Rearranging the terms, we get the acceleration (α) of the center of the hoop:

α = (mg * sin(θ)) / (2m) = (g * sin(θ)) / 2

For the hoop to roll without slipping, the static friction (f) must be equal to the torque acting on the hoop. The minimum coefficient of static friction (μmin) can be determined using the frictional force equation:

f = μmin * N

Since the normal force (N) equals mg * cos(θ), the frictional force can be written as:

f = μmin * mg * cos(θ)

Now, substituting f = m * α from equation (3) into this equation, we get:

m * α = μmin * mg * cos(θ)

Rearranging the terms, we find the minimum coefficient of static friction (μmin) needed for the hoop to roll without slipping:

μmin = α / (g * cos(θ)) = (g * sin(θ) / 2) / (g * cos(θ)) = tan(θ) / 2

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1) Calculate the change in entropy (AS) for the following processes: a) Heating 0.5 kg of solid ice from -100°C to 0°C. b) Freezing 0.5 kg of solid ice at 0°C to liquid water at 0°C. c) Cooling 0.5 kg of liquid water from 100°C to 0°C. d) Vaporizing 0.5 kg of liquid water at 100°C to steam at 100°C. e) Which one of the two phase transitions (freezing or vaporizing) has the larger magnitude in the change of entropy? Explain in terms of the statistical model of entropy. f) For each process a)-d) does the entropy of the environment change? If so, does it increase or decrease? How do you know?

Answers

the change in entropy is: A- ΔS = Q/T = (105,000 J) / (273 K) = 384.6 J/K b-  ΔS = -Q/T = (-166,750 J) / (273 K) = -611.2 J/K,c- ΔS = -Q/T = (-209,300 J) / (373 K) = -560.4 J/K, d)- ΔS = Q/T = (1,128,500 J) / (373 K) = 3023.5 J/K

a) To calculate the change in entropy when heating 0.5 kg of solid ice from -100°C to 0°C, we can use the equation:

ΔS = Q/T

where Q is the heat transferred to the system and T is the temperature at which the heat transfer occurs. Since the system is receiving heat, the sign of ΔS will be positive.

The heat required to raise the temperature of the ice from -100°C to 0°C can be calculated as:

Q = mcΔT

where m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.

Using the values given, we get:

Q = (0.5 kg) (2100 J/kg°C) (100°C) = 105,000 J

Therefore, the change in entropy is:

ΔS = Q/T = (105,000 J) / (273 K) = 384.6 J/K

b)  Q = (0.5 kg) (333,500 J/kg) = 166,750 J

Therefore, the change in entropy is:

ΔS = -Q/T = (-166,750 J) / (273 K) = -611.2 J/K

c) Q = (0.5 kg) (4186 J/kg°C) (100°C) = 209,300 J

Therefore, the change in entropy is:

ΔS = -Q/T = (-209,300 J) / (373 K) = -560.4 J/K

d) Q = (0.5 kg) (2,257,000 J/kg) = 1,128,500 J

Therefore, the change in entropy is:

ΔS = Q/T = (1,128,500 J) / (373 K) = 3023.5 J/K

e) The change in entropy for vaporizing water (3023.5 J/K) is larger

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sirius, the brightest star in the night sky, is actually a binary star system. sirius a is main-sequence star and sirius b is a white dwarf. nearly all the visible light we see from sirius comes from sirius a. but when we photograph the system with x-ray light, as shown here, sirius b is the brighter of the two stars. why?

Answers

Sirius, the brightest star in the night sky, is actually a binary star system. Sirius A is a main-sequence star, and Sirius B is a white dwarf. Nearly all the visible light we see from Sirius comes from Sirius A. But when we photograph the system with X-ray light, as shown , Sirius B is the brighter of the two stars because As a white dwarf, Sirius B is much hotter than Sirius A and thus emits more X-rays.

Sirius is a binary star system made up of two stars called Sirius A and Sirius B.  Sirius A is the larger and more massive star, while Sirius B is a smaller, denser, and hotter white dwarf. Sirius A is a main-sequence star, similar to our Sun, while Sirius B is the remnant of a star that has exhausted all of its nuclear fuel and has collapsed down to a dense core.

Although Sirius A is much brighter than Sirius B in visible light, as mentioned, Sirius B is actually the brighter of the two stars when observed in X-ray light. This is because as a white dwarf, Sirius B has a much smaller surface area than Sirius A, but is much hotter and therefore emits more energy per unit area.

The high temperature of Sirius B's surface causes it to emit a significant amount of X-ray radiation, which is why it appears brighter than Sirius A in X-ray images of the system.

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The probable question may be:

sirius, the brightest star in the night sky, is actually a binary star system. sirius a is main-sequence star and sirius b is a white dwarf. nearly all the visible light we see from sirius comes from sirius a. but when we photograph the system with x-ray light, as shown here, sirius b is the brighter of the two stars. why?

a coil has an inductance of 7.00 mh, and the current in it changes from 0.200 a to 1.50 a in a time interval of 0.150 s. find the magnitude of the average induced emf in the coil during this time interval.

Answers

A coil has an inductance of 7.00 mh, and the current in it changes from 0.200 a to 1.50 a in a time interval of 0.150 s. The magnitude of the average induced emf in the coil during this time interval is 0.72 V.

What is induced EMF, The induced EMF (electromotive force) is the voltage that is generated when a magnetic field passes through a conductor or is generated when a conductor moves through a magnetic field.

The formula for the magnitude of the average induced EMF in a coil is given by the following formula;

E = L (ΔI / Δt)

Where L is the inductance of the coil, ΔI is the change in current, and Δt is the time interval.

Substituting the given values,

L = 7.00 mH = 7.00 × 10-3 HΔI

= 1.50 A − 0.200 A

= 1.30 AΔt

= 0.150

sE = L (ΔI / Δt) = (7.00 × 10-3 H) (1.30 A / 0.150 s)E

= 0.072 V

therefore, The magnitude of the average induced EMF in the coil during this time interval is 0.72 V.

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If a piece of COLD metal is dropped into a sample of LUKEWARM water, how will the temperature of the water change?

!See how COLD metal is capitalized!

Answers

The water molecules move less quickly because some of the energy from the water was used to accelerate the metal. The water's temperature drops as a result of this.

What occurs if you drop a hot piece of metal into cold water?

The metal will eventually chill while the water warms up. The temperatures of the two items will eventually be equal. When this occurs, it is said that they are in thermal balance with one another.

What occurs if you add cold water to hot water?

Because heat moves from higher temperature to lower temperature when hot water and frigid water are combined, the mixture reaches an intermediate temperature. when hot water is followed by frigid water.

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the electron density between is and 2s orbital is very high at the nucleus, but the radial probability drops ot zero at the nucleus. explain why this is true

Answers

The electron density between the 1s and 2s orbitals is indeed high at the nucleus, but the radial probability density drops to zero at the nucleus. This phenomenon is known as the "electron density paradox".

To understand why this is true, we need to consider the wave nature of electrons in an atom. According to the Schrödinger equation, the wave function of an electron is a probability amplitude that describes the probability of finding an electron in a particular location in space. The radial probability density is the probability of finding an electron at a particular distance from the nucleus. In the case of the 1s orbital, the electron has a high probability of being found at the nucleus because the wave function has a large amplitude there. However, the probability density drops to zero at the nucleus because the electron is a wave that must obey the laws of quantum mechanics, including the Heisenberg uncertainty principle. This principle states that it is impossible to determine both the position and momentum of an electron with absolute precision.

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for a converging lens with two curved surfaces, the radius of curvature for both surfaces is 10 cm. if the focal length is 10 cm, what must the index of refraction be?

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The index of refraction must be 2, which is not physically possible, for a converging lens with two curved surfaces.

The central length of a combining focal point with two bended surfaces is given by the recipe 1/f = (n-1)[(1/R1) - (1/R2)], where f is the central length, n is the record of refraction, R1 and R2 are the radii of shape of the two surfaces. For this situation, the two surfaces have a span of shape of 10 cm, and the central length is likewise 10 cm. Accordingly, we can substitute these qualities into the equation and tackle for n:

1/10 = (n-1)[(1/10) - (1/10)]

n-1 = 1

n = 2

Hence, the record of refraction should be 2. It's quite significant that this worth isn't truly workable for a material to have, as the refractive file of any material should be more noteworthy than or equivalent to 1. Nonetheless, this outcome recommends that the focal point being referred to has an uncommon shape or material properties.

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if a balloon was inflated with air to a volume of 6.0 l under the conditions in mongolia and then rapidly transported to the conditions described for the typhoon in the pacific, what would the volume of the balloon be under these new conditions, assuming no change in temperature?

Answers

The initial volume of the balloon when it was inflated under the conditions in Mongolia is 6.0 L. The volume of the balloon under the new conditions described for the typhoon in the Pacific is 5.6 L, assuming no change in temperature.

The conditions in Mongolia are not specified, but it is clear that the conditions in the Pacific are different. The question asks for the volume of the balloon under the new conditions described for the typhoon in the Pacific. Since there is no change in temperature, we can use the ideal gas law, which is PV = nRT.

The volume of the balloon will change due to the change in pressure. The pressure will increase under the new conditions in the Pacific. Therefore, the volume will decrease. The formula for this is:

P1V1 = P2V2

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

We can use this formula to find the final volume of the balloon under the new conditions in the Pacific.P1 = initial pressure = unknownV1 = 6.0 LP2 = final pressure = unknownV2 = final volume

We know that the volume of the balloon will decrease, so V2 < V1. We can solve for V2 as follows:P1V1 = P2V2V2 = (P1V1)/P2Now we need to find the pressure in Mongolia and the pressure in the Pacific.

We can use the average sea level pressure for each location as a reference. The average sea level pressure in Mongolia is about 1000 hPa. The average sea level pressure in the Pacific is about 1013 hPa. We can use these values as estimates for P1 and P2, respectively.

P1 = 1000 hPaV1 = 6.0 LP2 = 1013 hPaV2 = (P1V1)/P2V2 = (1000 hPa * 6.0 L)/1013 hPaV2 = 5.6 L

Therefore, the volume of the balloon under the new conditions described for the typhoon in the Pacific is 5.6 L, assuming no change in temperature.

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what value in a raster surface representing travel impedance would be treated as absolutely impassable?

Answers

In a raster surface representing travel impedance, a value of 0 would be treated as absolutely impassable. What is a raster surface.

A raster surface is a type of data that uses a grid of uniformly spaced cells or pixels to represent a continuous surface. The elevation of the land, the temperature of the ocean's surface, and the intensity of reflected light are examples of continuous surfaces. Raster surfaces are frequently used in Geographic Information Systems (GIS) to represent spatial phenomena such as elevation, temperature, rainfall, and so on. This sort of data is utilized to make maps that are extremely helpful in a variety of fields. What is Travel Impedance. In GIS, travel impedance is the resistance that limits or prevents movement between two points. This resistance could be in the form of terrain, traffic, weather, or other environmental factors that could limit travel. Travel impedance is commonly utilized to predict or calculate the shortest route or travel time between two points in a spatial data environment. What is the value in a raster surface that represents travel impedance that is absolutely impassable. A value of 0 in a raster surface that represents travel impedance is treated as absolutely impassable. This implies that the location is either inaccessible or a barrier, and travel through that location is not feasible.
In a raster surface representing travel impedance, a value of infinity or a very high value would be treated as absolutely impassable.

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Describe how the location of the oxygen molecules will change after the divider swings open.

Answers

Find the freezing point of the material and contrast it to that of other metals. Measure the metal's length and contrast it with the lengths of other metals. Also, contrast the metal's shape with those of other metals.

Which of the following best describes how heat energy flows in the concentrated solar layout?

The air within the solar cooker receives energy from the water through the edges of the cook pot. The air outside solar cooker receives energy that is transferred from the gas within the solar cooker through sidewalls of the solar cooker.

What kind of energy is transferred in a solar cooker?

Radiation is utilized to bake the s'mores in the hot cooker that is created in this exercise. When heat is transported without the presence of Sun and s'mores were in close proximity to one another  Electromagnetic waves that are passing through the air convey the heat.

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a person having a mass of 80 kg sits on the edge of a horizontal rotating platform, 1.9 m from the center of the platform, and has a tangential speed of 6.0 m/s . calculate the angular momentum of the person.

Answers

Answer:

L = 912 kg*m^2/s

Explanation:

L = mvr

L = 80kg(6m/s)(1.9m)

L = 912

The angular momentum of the person is 912 kg⋅m²/s.

The angular momentum (L) of an object is given by the product of its moment of inertia (I) and its angular velocity (ω). The moment of inertia of a point mass rotating about an axis at a distance r from the center is given by I = mr², where m is the mass of the object.

In this case, the person is sitting on the edge of a rotating platform at a distance of 1.9 m from the center. Therefore, the moment of inertia of the person-platform system is I = mr² = 80 kg × (1.9 m)² = 288.8 kg⋅m².

The tangential speed (v) of the person is given as 6.0 m/s. The angular velocity (ω) can be calculated as ω = v/r, where r is the distance from the center. Therefore, ω = 6.0 m/s ÷ 1.9 m = 3.16 rad/s.

Now, we can calculate the angular momentum of the person as L = Iω = (288.8 kg⋅m²) × (3.16 rad/s) = 912 kg⋅m²/s.

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A cyclist is travelling along a straight path at 5.5 m/s.
The cyclist accelerates uniformly to a speed of 11.0 m/s.
The acceleration of the cyclist is 1.2 m/s².
Calculate the distance travelled by the cyclist during this acceleration.
Give your answer correct to 2 significant figures.
Use the equation: v² - u² = 2 x a x X

Answers

Answer:

37.5 meters

Explanation:

We can use the equation:

v² - u² = 2ax

where v is the final velocity, u is the initial velocity, a is the acceleration, and x is the distance travelled during the acceleration.

We are given:

u = 5.5 m/s

v = 11.0 m/s

a = 1.2 m/s²

Substituting these values into the equation, we get:

11.0² - 5.5² = 2 × 1.2 × x

Simplifying the equation, we get:

120.25 - 30.25 = 2.4x

90 = 2.4x

x = 37.5 meters

Therefore, the distance traveled by the cyclist during the acceleration is 37.5 meters (to 2 significant figures).

How does the direction of acceleration change when the masses on an Atwood's machine are switched, and how can this be explained using Newton's laws of motion?

Answers

The acceleration of the system still depends only on the difference in mass between the two masses, but the direction of the net force has changed, so the direction of acceleration has also changed.

When the masses are different, the system accelerates in the direction of the heavier mass. According to Newton's second law, the net force acting on each mass is equal to its mass times its acceleration:

F1 = m1a

F2 = m2a

where F1 and F2 are the tensions in the rope on either side of the pulley. Since the rope is inextensible and the pulley is ideal, the tension in the rope is the same on both sides, so F1 = F2 = T, where T is the tension in the rope.

Substituting T for F1 and F2 in the above equations and subtracting the second equation from the first, we get:

m1a - m2a = T - T

(m1 - m2)a = 0

Since the tension cancels out, the acceleration of the system depends only on the difference in mass between the two masses.

If the masses are switched, then the direction of the net force changes, and the system accelerates in the direction of the new heavier mass. The net force acting on each mass is still given by the above equations, but the masses are switched, so m1 is now the smaller mass, and m2 is the larger mass.

Substituting T for F1 and F2 in the above equations and subtracting the first equation from the second, we get:

m2a - m1a = T - T

(m2 - m1)a = 0

Since the tension cancels out, the acceleration of the system still depends only on the difference in mass between the two masses.

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A submarine has a mass of 5 tonnes. State the buoyancy required for it to float stationary in the water

Answers

Answer: The buoyancy required for it to float stationary in the water is 49000

Explanation:

A 0.10-kilogram piece of modeling clay is tossed at a motionless 0.10-kilogram block of wood and sticks. The block slides across a frictionless table at 15 m/s.
a. At what speed was the clay tossed?
b. The clay is replaced with a “bouncy” ball tossed with the same speed. The bouncy ball rebounds from
the wooden block at a speed of 10 m/s. What effect does this have on the wooden block? Why?

Answers

The clay was tossed at a speed of 15 m/s. The system's overall momentum is still preserved, but its total kinetic energy is not.

What does conservation of momentum?

In an isolated system, when two objects contact, the total momentum before and after the collision is equal, according to the law of conservation of momentum. This is because the momentum that is lost by one object and acquired by another is equivalent.

a. we can use the conservation of momentum equation:

m1v1 + m2v2 = (m1 + m2)v

m1 and v1 are the mass and starting velocity of the clay, m2 and v2 are the mass and initial velocity of the wooden block (which is immobile), respectively.

Substitute in the given values,

(0.10 kg) v1 + (0.10 kg) (0 m/s) = (0.10 kg + 0.10 kg) (15 m/s)

Solving for v1,

v1 = 15 m/s

b. The wooden block experiences force in the opposite direction of the motion as the bouncy ball bounces off of it. The block finally slows down and stops as a result of this. The wooden block is affected by the bouncing ball in such a way that it receives some of the ball's kinetic energy and moves in the opposite direction of the ball. The wooden block slows down and comes to a stop, while the ball continues to move because it bounces back and retains some of its kinetic energy.

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) If you were on Mars where the gravitational constant is 3.71 m/s2, how far would a rock
dropped in Valles Marineris (a system of canyons on Mars up to 7,000 m deep) fall after
9 seconds?

Answers

Therefore, a rock dropped in Valles Marineris on Mars would fall approximately 150.255 meters after 9 seconds.

On Mars, what shaped the Valles Marineris?

The majority of scientists concur that Valles Marineris is a sizable tectonic "crack" that formed in the Martian crust as the planet cooled, was influenced by the rising crust in the Tharsis area to the west, and then widened by erosional forces.

After 9 seconds, an object dumped on Mars would have travelled the distance indicated by the following formula:

d = 0.5 * g * t²

When we enter the specified numbers into the formula, we obtain:

d = 0.5 * 3.71 m/s² * (9 s)²

d = 0.5 * 3.71 m/s² * 81 s²

d = 150.255 m

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what is the potential difference between yi =−7cm and yf =7cm in the uniform electric field e⃗ =(20,000i^−75,000j^)v/m?

Answers

The potential difference between the points at yi = -7cm and yf = 7cm in the given electric field is 1,050,000 V.

The electric field is given as E⃗ = (20,000i^ - 75,000j^) V/m. This means that the electric field is only in the y-direction, and its magnitude is constant throughout.

To find the potential difference between two points in the electric field, we can use the following formula:

ΔV = -∫E⃗ · dl⃗

where ΔV is the potential difference, E⃗ is the electric field, and dl⃗ is an infinitesimal displacement along the path between the two points. The negative sign arises because the electric field is acting in the opposite direction to the direction of motion along the path.

In this case, we can assume a path that moves from the point at yi = -7cm to the point at yf = 7cm, both of which lie on the y-axis. Since the electric field is only in the y-direction, we can assume that the path is also along the y-axis. Therefore, we can simplify the formula to:

ΔV = -∫E dy

where the integral is taken from yi = -7cm to yf = 7cm.

Substituting the electric field E⃗ and integrating, we get:

ΔV = -∫E dy = -∫(-75,000) dy (since the y-component of the electric field is -75,000 V/m)

ΔV = -[-75,000y]yi=-7cm to yf=7cm

ΔV = -[-75,000(7cm - (-7cm))] = 1,050,000 V

Therefore, the potential difference between the points at yi = -7cm and yf = 7cm in the given electric field is 1,050,000 V.

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1. for a classical wave, which of the following is true: a. frequency is proportional to velocity b. wavelength is proportional to frequency c. energy is proportional to the amplitude d. energy is proportional to wavelength

Answers

For a classical wave, the following statements are true: b,c and the statements a and d are false.

a. Frequency is proportional to velocity: This statement is not true. Frequency and velocity are independent properties of a wave. The frequency of a wave refers to the number of oscillations or cycles per unit time, while the velocity of a wave refers to the speed at which the wave propagates.

b. Wavelength is proportional to frequency: This statement is partially true. In a classical wave, the wavelength is inversely proportional to the frequency. This means that as the frequency increases, the wavelength decreases, and vice versa.

c. Energy is proportional to the amplitude: This statement is true. The amplitude of a wave refers to the maximum displacement of the wave from its equilibrium position. The energy of a wave is proportional to the square of its amplitude. This means that as the amplitude increases, the energy carried by the wave also increases.

d. Energy is proportional to wavelength: This statement is not true. The energy of a classical wave is proportional to its frequency, not its wavelength. This is expressed by the formula E = hf, where E is the energy of the wave, h is Planck's constant, and f is the frequency of the wave.

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transformer a has a primary coil with 400 turns and a secondary coil with 200 turns. transformer b has a primary coil with 400 turns and a secondary coil with 800 turns. the same current and voltage are delivered to the primary coil of both transformers. the secondary coils are connected to identical circuits. how does the power delivered by secondary coil a compare to the power delivered by secondary coil b?

Answers

The power delivered by both secondary coils A and B is the same (P_A = P_B), the comparison shows that the power delivered by secondary coil A is equal to the power delivered by secondary coil B.

Transformer A has a primary coil with 400 turns and a secondary coil with 200 turns. Transformer B has a primary coil with 400 turns and a secondary coil with 800 turns. The same current and voltage are delivered to the primary coil of both transformers, and the secondary coils are connected to identical circuits.

Step 1: Calculate the turns ratio for each transformer.
Transformer A:

Turns ratio (N) =

[tex]= N_{secondary} / N_{primary[/tex]

= 200 turns / 400 turns

= 0.5
Transformer B:

Turns ratio (N)

= [tex]N_{secondary} / N_{primary[/tex]

= 800 turns / 400 turns

= 2

Step 2: Determine the voltage and current ratios.
For transformers, the voltage ratio (V) and the current ratio (I) are inversely proportional to the turns ratio.
Transformer A:

Voltage ratio (V_A)

[tex]= N_A = 0.5,[/tex]

Current ratio (I_A)

[tex]= 1 / N_A = 2[/tex]
Transformer B:

Voltage ratio (V_B)

[tex]= N_B = 2,[/tex]

Current ratio (I_B)

[tex]= 1 / N_B = 0.5[/tex]

Step 3: Calculate the power delivered by the secondary coils.
Power (P) = Voltage (V) × Current (I)
For Transformer A:

[tex]P_A = V_A \times I_A \\= 0.5 \times 2 = 1[/tex](relative value)
For Transformer B:

[tex]P_B = V_B \times I_B \\= 2 \times 0.5 = 1[/tex](relative value)

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building a supermassive black hole. the black hole in the galaxy m106 has a mass of about 36 million solar masses. let us assume that most of that mass flowed into the black hole through an accretion disk that radiated 10% of the mass-energy ( ) passing through it. in that case, what would be the total amount of energy radiated by the accretion disk during the history of the black hole? what would be the average luminosity of the accretion disk, if it continuously radiated that energy over a period of 10 billion years? how does that average luminosity compare with the luminosity of the milky way?

Answers

The total amount of energy radiated by the accretion disk during the history of the black hole is approximately  6.44 x 10^53 Joules. The average luminosity of the accretion disk  is approximately 2.04 x 10^37 Watts.

The average luminosity of the accretion disk is slightly lower than the luminosity of the Milky Way.

To determine the total amount of energy radiated by the accretion disk during the history of the black hole, we can use the mass-energy equivalence formula, E=mc^2, where E is the energy, m is the mass, and c is the speed of light (approx. 3x10^8 m/s).

First, we find 10% of the black hole's mass: 0.1 x 36 million solar masses = 3.6 million solar masses. Then, we convert this mass to kilograms: 3.6 million solar masses x (1.989 x 10^30 kg/solar mass) ≈ 7.16 x 10^36 kg.

Next, we calculate the energy: E = (7.16 x 10^36 kg) x (3 x 10^8 m/s)^2 ≈ 6.44 x 10^53 Joules.
To find the average luminosity of the accretion disk, we divide the total energy by the time it radiated that energy: 6.44 x 10^53 Joules / (10 billion years x 3.1536 x 10^7 s/year) ≈ 2.04 x 10^37 Watts.

Comparing the average luminosity of the accretion disk to the luminosity of the Milky Way, which is around 3 x 10^37 Watts, we see that the average luminosity of the accretion disk is slightly lower than the luminosity of the Milky Way.

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what is the efficiency (in percent) of a subject on a treadmill who puts out work at the rate of 117 w while consuming oxygen at the rate of 2.12 l/min?

Answers

The efficiency of a subject on a treadmill can be calculated using the ratio of work output over work input. In this case, the work output is 117 W and the work input is 2.12 L/min of Oxygen.

The efficiency can be calculated as (117 W) / (2.12 L/min) * 100, which is approximately 55%. This means that 55% of the energy the subject is consuming is being used for work output and the remaining 45% is being used for other purposes such as heat loss or respiration. In other words, the subject is using 55% of their energy efficiently to produce work. In conclusion, the efficiency of the subject on a treadmill is 55%.

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g the voltage across a capacitor is 12 volts. if its capacitance is 3 farads, how much charge (in coulombs) is stored in it?

Answers

The charge stored in the capacitor is 36 coulombs.

When the voltage across a capacitor is 12 volts and its capacitance is 3 farads, the amount of charge stored in it is 36 coulombs.

What is voltage?

The voltage is the electric potential difference between two points in a circuit, or it may be the driving force that causes current to flow. The charge q stored in a capacitor with a capacitance C when a voltage V is applied is given by: q = CV, where q is the charge stored in the capacitor, C is the capacitance of the capacitor, and V is the voltage applied to the capacitor. A 3-farad capacitor that is charged to 12 volts stores 36 coulombs of charge. Therefore, 36 coulombs of charge is stored in it.

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please help `\_('-')_/`

Answers

The net force acting on an object is the vector sum of all the forces acting on it. Since the resulting net force is -10 N, we can deduce that the forces acting on the object are in opposite directions. Therefore, we need to choose the pair of forces whose vector sum is -10 N.

Option C (-63 N and 73 N) satisfies this condition because -63 N + 73 N = 10 N in the opposite direction, which matches the resulting net force of -10 N.

Therefore, the most likely represent the forces acting on the object are -63 N and 73 N. Answer: C.
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