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Valentin [98]
3 years ago
9

A 2000kg car experiences a braking force of 10000N and skids to a stop in 6 seconds. The speed of the car just before the brakes

were applied was
A) 45 m/s
B) 15 m/s
C) 1.2 m/s
D) 30 m/s
E) none of these
So far I got a= 5m/s^2 and u= 30m/s but I think the answer is 45 m/s but I do not know where to go from there.
Physics
1 answer:
mylen [45]3 years ago
4 0

Answer:

D. 30m/s

Explanation:

According to the equation of motion

v=u +at

v = Final velocity

u = initial velocity

a = acceleration

t = time taken

Since we don't know the acceleration of the body, we will find its acceleration using newtons first law of motion F = ma

10000 =2000a

a = 5m/s²

initial velocity = 0m/s since the body starts from rest

v is what we are looking for

t = 6seconds

Substituting this datas in the formula

v = 0+ 5(6)

v = 30m/s (D)

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"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much
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The number of bright fringes per unit width on the screen is, x=\dfrac{\lambda D}{d}      

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6 0
3 years ago
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
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
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