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Colt1911 [192]
2 years ago
14

In the image below, which wire loop is experiencing the greatest magnetic flux?

Physics
2 answers:
iogann1982 [59]2 years ago
7 0

Answer:

image D is experiencing the greatest magnetic flux

saveliy_v [14]2 years ago
6 0
Image D is the correct answer
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A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at
Stells [14]
The correct answer is <span>3) K_f =  \frac{1}{2}mv_0^2 + mgh.
</span>
In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span>E=U_i+K_i=mgh +  \frac{1}{2}mv_0^2
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy: 
</span>E=K_f<span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
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2 years ago
When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

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2 years ago
The population in the United States in 2015 was 321 million people. It is projected to increase to 438 million people by the yea
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It increases by 35% ......

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2 years ago
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In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you
olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

\Delta y = 5 - 1.6 = 3.40 m

now we will have displacement in horizontal direction

\Delta x = 14 m

now we know that

v_x = vcos54

v_y = vsin54

now we will have

\Delta x = v_x t

14 = (vcos54)t

also for y direction

\Delta y = v_y t + \frac{1}{2}at^2

3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2

now from the two equations we will have

3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2

3.40 = 14 tan54 - 4.9 t^2

3.40 = 19.3 - 4.9 t^2

t = 1.8 s

now from above equations

14 = vcos54 (1.8)

v = 13.2 m/s

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

\Delta y = -1.60 m

now we will have

-1.60 = v_y t + \frac{1}{2}at^2

-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2

4.9 t^2 - 10.7t - 1.60 = 0

t = 2.3 s

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

t_2 = t - t_1

t_2 = 2.3 - 1.8 = 0.5 s

now the displacement on the other side is given as

\Delta x = (vcos54) t_2

\Delta x = (13.2 cos54)(0.5)

\Delta x = 3.88 m

4 0
2 years ago
First find ∮RB⃗ ⋅dl⃗ , the line integral of B⃗ around a loop of radius R located just outside the left capacitor plate. This can
Allisa [31]

Answer:

the expression of current in the loop enclosed to the left of the capacitor plate is

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Explanation:

As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path

So we will have

\int B. dL = \mu_0 I(t)

so we have

I(t) = \frac{1}{\mu_0}\int B. dL

so above is the expression of current in the loop enclosed to the left of the capacitor plate

5 0
2 years ago
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