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scoundrel [369]
3 years ago
8

During the launch from a board, a diver's angular speed about her center of mass changes from zero to 8.10 rad/s in 240 ms. Her

rotational inertia about her center of mass is 11.6 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?
Physics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

Part a)

\alpha = 33.75 rad/s^2

Part b)

\tau = 391.5 Nm

Explanation:

Part a)

Average angular acceleration is given as rate of change in angular speed

so it is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{8.10 - 0}{240\times 10^{-3}}

\alpha = 33.75 rad/s^2

Part b)

average external torque is given as

\tau = I\alpha

here we know that

I = 11.6 kg m^2

\tau = 11.6 \times 33.75

\tau = 391.5 Nm

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Answer:

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Direction: negative

Explanation:

From Newton's law of motion, we know that;

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Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

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B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

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Select the situation for which the torque is the smallest.
alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

a)

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

<u>τ = 4900 N.m</u>

<u></u>

b)

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

<u>τ = 4900 N.m</u>

<u></u>

c)

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

<u>τ = 4900 N.m</u>

<u></u>

Hence, the correct answer will be:

<u>e. The torque is the same for all cases.</u>

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