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Alexxx [7]
3 years ago
6

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

in 6.73s. what is the magnitiude of the girls average acceleration.
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

a = 0.53 m/s^2

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

\omega = 2\pi f

\omega = 2\pi ( \frac{20}{60})

\omega = 2.10 rad/s

so final tangential speed is given as

v = r\omega

v = 1.71 (2.10) = 3.58 m/s

now average acceleration of the girl is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{3.58 - 0}{6.73}

a = 0.53 m/s^2

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
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Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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Answer: D

Explanation:

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a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi
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The radius of the sphere in meters is ,r = 10\sqrt{5} -20

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

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