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Effectus [21]
3 years ago
8

A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to

South. A coil of resistance R is connected to a battery which supplies a potential difference of V across the coil and is suspended such that a section of the coil of length L meters lies between it with current running from East to West. The mass balance measures a mass of M.
1. What is the measured change in mass due to the effect of Fmag?
2. What is the total measured mass of the magnets? Keep in mind the effect of Newton's third law.

Physics
1 answer:
rjkz [21]3 years ago
7 0

Answer:

Explanation:

FIND THE SOLUTION BELOW

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When the acceleration of a mass on a spring is zero, the velocity is at a
Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

F=kx

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

F=ma

where

m is the mass

a is the acceleration

So we can write the equation as

ma=kx (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const. (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a  maximum

When the velocity of a mass on a spring is zero, the acceleration is at a  maximum

6 0
3 years ago
You are not harmed by contact with a charged metal ball, even though its voltage may be very high. Is the reason similar to why
Jobisdone [24]

Answer:

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Explanation:

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6 0
3 years ago
The energy due to Earth pulling down on an object is called ___.
Slav-nsk [51]

Gravity is the energy due to Earth pulling down on an object.

6 0
4 years ago
During a shock which lasts 10ms (0.01s), the voltage difference between the electrodes (so the drop in potential across all thre
natta225 [31]

Answer:

20000 W

Explanation:

Power: This can be defined as the rate at which energy is dissipated or used. The S.I unit of power is Watt(W).

The expression of power is given as,

P = E/t.............................. Equation 1

Where P = power, E = Energy, t = time.

Given: E = 200 J, t = 0.01 s

Substitute into equation 1

P = 200/0.01

P = 20000 W.

Hence the average power = 20000 W

5 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
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