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vfiekz [6]
4 years ago
4

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in time t, th

rough what angle did it rotate in the time ½t?
a. 4θ
b. ¼θ
c. ½θ
d. 2θ
e. θ
Physics
1 answer:
Blizzard [7]4 years ago
4 0

Answer:

Angular displacement will be \frac{1}{4}\Theta

So option (b) will be the correct option

Explanation:

We have given that firstly object is at rest

So \omega _i=0rad/sec

From law of motion we know that angular displacement is given by

\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2

Now angular displacement by the object in \frac{t}{2}sec

\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta

So option (b) will be the correct option

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Ludmilka [50]

Answer: 28.699

Explanation:

KE=1/2(mv²)

1800=1/2m (125.44)

14.349=2m

m=28.699

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Can you ever hold your Moon sphere in the light and have it be more or less than half-lit?
Anettt [7]
No if in the light it would be always half lit
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Each student aligns the edge of the object with the left edge of the ruler and records a length of 4.6 cm. Which statement descr
TEA [102]
If they align the object at the very edge of the ruler, the measurement in inaccurate. They would have to start the measurement at 0 to get an accurate measurement. When starting at the very edge, you are not taking into account the little bit of the ruler and it between the edge of the ruler and the 0 marker.
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3 0
3 years ago
A box weights 50kg applies friction force of 125N. You apply a force of 150N going in the right direction. What is the total net
Pachacha [2.7K]

Answer:

Net force  = 25N  and in the right direction

Explanation:

mass of box, m = 50kg

Force of block = 150N

Friction force = 125N

Net force  = Force Applied - Friction force =  F - F_f

150 - 125 = 25N

the force applied to the block is greater than the friction force. there the force applied will overcome the friction force and move in the right direction of the force applied

4 0
4 years ago
horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, wa
VARVARA [1.3K]

Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform \omega = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle m_p = 20.5 kg

Your mass  m' = 73.5 kg

speed v = 1.05 m/s with respect to the platform

V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525  \ m /s  \\  \\r = \frac{R}{2}

r = 0.955

Mass of the mutt m_m = 18.5 kg

r' = \frac{3}{4} \ R

Your angular momentum is calculated as:

Your angular velocity relative to the platform is \omega' = \frac{v}{R}  = \frac{1.05}{1.91 } = 0.5497 \ rad/s

Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s

I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2

L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s

For poodle :

Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \  rad/s

Actual \omega_p = \omega - \omega' =  0.919 -0.550 = 0.369 \ rad/s

I_p = m_p(\frac{R}{2} )^2  = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2

L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \  kgm/s

I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \  kgm^2

L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s

Disk I = \frac{mr^2}{2} =  \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2

L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s

Total angular momentum of system is:

L = L_D +L_Y+L_P+L_M

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

6 0
4 years ago
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