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Schach [20]
3 years ago
8

Consider a car driving around a circular track. Once started, the driver maintains a speed of 45 m/h. Compare the driver's veloc

ity with his acceleration in this case.
A) The car is accelerating because it is changing direction but the velocity is constant.
B) The car is traveling at a constant rate of speed so it is not accelerating and the velocity does not change.
C) Acceleration is the change in velocity so the car is experiencing a change in velocity and it is accelerating.
D) The car is traveling at a constant rate of speed so it is not accelerating but the velocity changes because the direction the car travels is constantly changing.
Physics
1 answer:
DENIUS [597]3 years ago
8 0
And sbdjdjdjfjfjfbfbfjf

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Sunny_sXe [5.5K]

Answer:

P V = n R T      ideal gas equation

P2 V2 / P1 V1 = T2 / T1    

V2 / V1 = T2 / T1 * P1 / P2 = T2 P1 / (T1 P2)

V2 / V1 = (1.17 T1) / T1 * (P1 / .22 P1)      assuming absolute temp as 1.17 P1

V2 / V1 = 1.17 / .22 = 5.32

V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 4/3 pi D^3 / 8 = pi D^3 / 6

V2 / V1 = D2^3 / D1^3

D2 = (V2 / V1 * D1^3)^1/3

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7 0
3 years ago
Large radio telescopes, like the one in Arecibo, Puerto Rico, can detect extremely weak signals. Suppose one radio telescope is
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Answer:

Explanation:

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4 0
3 years ago
12. A 20-newton cart is lifted up 0.40 meters to the top of a ramp.
Hitman42 [59]

Answer:

8 joules

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8 0
3 years ago
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A water wave passes by a floating leaf that is made to oscillate up and down two complete cycles each second, which means that t
lina2011 [118]

Answer:

2 Hz.

Explanation:

Frequency is simply defined as the number of appearances of a periodic event occurring per time. It is usually measured in cycles/second.

Now, in this question, we are told that there are 2 cycles for each second.

Thus, we can say that the frequency is 2 cycles/1 s = 2 Hz.

7 0
3 years ago
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

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At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
4 years ago
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