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Rashid [163]
3 years ago
12

A disk is free to rotate about an axis perpendicular to the disk through its center. If the disk starts from rest and accelerate

s uniformly at the rate of 2 radians/S2 for 4 s, its angular displacement during this time is
(A) 6 radians
(B) 12 radians
(C) 16 radians
(D) 24 radians
(E) 48 radians
Physics
1 answer:
garri49 [273]3 years ago
3 0

Answer:

(C) 16 radians

Explanation:

The angular displacement is given by the following equation:

\Delta \theta=\omega_i t+\frac{1}{2}\alpha t^2

Here

\Delta \theta Is the angular displacement of the body at the indicated time (t).

\omega_i Is the angular velocity of the body at the initial moment.

\alpha Is the angular acceleration of the body.

The disk starts from rest, so \omega_i=0

Replacing the given values:

\Delta \theta=\frac{1}{2}(2\frac{radians}{s^2})(4s)^2\\\Delta \theta=16 radians

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KE= 3.7 J - answer
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A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the
harina [27]

Answer:

R=2.78\ \Omega

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

P=I^2R

Put all the values to find R.

R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega

So, the resistance is equal to 2.78\ \Omega.

7 0
2 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
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