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Afina-wow [57]
3 years ago
12

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a

magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

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A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola
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C I’m pretty sure!!!!!
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A car (mass of 830 kg) is sitting on a car lift in a shop (neglect the mass of the lift itself). While the car is being lifted u
OLEGan [10]
There are two force acting on an object that is being lifted. (1) the weight of the car, (2) the upward force. The difference of these force should be equal to the product of the mass and the acceleration. (This is the content of Newton's 2nd Law of Motion). If we let the lifting force be F,
                       F - (830)(9.8) = (830)(3.8)
The value of F from the equation is 11288 N. 
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3 years ago
Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c
masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us  find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

7 0
3 years ago
A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
3 years ago
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