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kati45 [8]
3 years ago
12

A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the

ball as it rolls the 10 meters? 1) 2.5m/s 2) 10m/s 3) 12.5m/s 4) 15m/s (Please show calculation)
Physics
1 answer:
antoniya [11.8K]3 years ago
6 0
Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

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Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal
SpyIntel [72]

Answer:

x=1.75m

Explanation:

From the exercise we have that

y_{o}=1.3m\\v_{o}=3m/s, \beta  =37\\

<em><u>To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first </u></em>

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0m

0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}

Solving for t

t=-0.36 s or t=0.73s

Since the <u>time</u> can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

x=v_{ox}t=3cos(37)(0.73)=1.75m

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What is the difference in PE
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Answer:

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Explanation:

Info given:

mass adult: 82.5kg

mass child: 14.7kg

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B. Gamma rays
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