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3 years ago

A car accelerates from rest for 8.0 s, and reaches a speed of 66 m/s. How far will the

Physics

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Answer: C

Explanation:

Find the acceleration using this kinematic equation:

$v_f=v_i+at\\66=0+a*8\\a=8.25m/s^2$

Now use this kinematic equation to find the displacement:

$v_f^2=v_i^2+2a*d\\66^2=0^2+2(8.25)d\\4356=16.5d\\d=264m$

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The critical angle between a medium and air is 430. What is the speed of light in that medium?

vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

= 1/sin(430)

= 1.06

then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

= (3×10^8)/(1.06)

= 281907786.2 m/s

Therefore, the speed of light is that medium is 281907786.2 m/s.

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2 years ago

A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate

Romashka [77]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

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2 years ago

Please help! 30Points!! An ultrasound machine uses waves to create images. The machine uses sound waves and which wave interacti

Gnesinka [82]

The correct answer should be C

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3 years ago

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A bungee jumper launches herself off a bridge, how would you describe her motion?

Blizzard [7]

It would depend on how she jumped off but based on it sounds it would be a curving motion

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2 years ago

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The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago