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3 years ago

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Consider a solenoid of length L, N windings, and radius b (L is much longer than b A current I). is flowing through the wire. If

the length of the solenoid became twice as long (2L), and all other quantities remained the same, the magnetic field inside the solenoid would _________.

1 answer:

wolverine [178]3 years ago

5 0

Answer:

The magnetic field inside the solenoid would decrease by a factor of 2.

Explanation:

The magnetic field, B, of a solenoid of length L, N windings, and radius b with a current, I, flowing through it is given as:

B = (N * r * I) / L

If the length of the solenoid is doubled, 2L,the magnetic field becomes:

B2 = (N * r * I) / 2L

B2 = ½ B

The magnetic field will decrease by a factor of 2.

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A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio

xxTIMURxx [149]

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

$v^2-u^2 = 2as$

substitute the values and solve for v

$v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s$

b) Use conservation of momentum:

Initial momentum of the system = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

3 0

3 years ago

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$

We also know the total time of flight,

t = 7.82 s

Therefore, we can find the horizontal distance travelled by the sack:

$d=v_x t = (60)(7.82)=469.2 m$

So, the sack will land 469.2 m from the original point below the airplane.

Learn more about free fall and projectile motion:

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#LearnwithBrainly

7 0

3 years ago

If you double the distance between you and the center of Earth, what happens to the strength of the gravitational field you expe

Lostsunrise [7]

Answer:

The strength of gravity decreases.

An example of that would be if you were in space; you float around because there's no gravity.

7 0

3 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$

So the acceleration of car due to friction is given as

$F_{net} = - \mu mg$

$a = \frac{F_{net}}{m}$

$a = -\mu g$

now it is given that

$\mu = 0.868$

$g = 9.81 m/s^2$

so here we have

$a = -0.868 * 9.81$

$a = -8.52 m/s^2$

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0

4 years ago