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lana [24]
3 years ago
8

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

does it take to accelerate from 91.0 km/hr to 104 km/hr?
Physics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t=\dfrac{v-u}{a}

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

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In a concave mirror parallel rays falling on it convergs at
ella [17]

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1) In a concave mirror parallel rays falling on it converges at F and 2F.

Explanation:

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Thus the location of converging point in concave mirrors will be based on the position or distance of object from the mirror. If the object distance is very far from the twice the focal length distance of mirror, then the converging point will be the focal point or F. And if the object is placed slightly greater than twice the distance of focal point, then the image will be obtained at 2F. But the parallel beams will be converging at F and 2F.

5 0
3 years ago
1)the car's engine power is 44000W. Explain this number in a physical sense
Ratling [72]

Answer:

1) It expresses the rate (top speed) at which it can move with time.

2) P = 20 W

3) h = 18 km

Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = \frac{Energy(or workdone)}{Time}

i.e P = \frac{E}{t}

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

    t = 20 s

h = 100m

g = 10 m/s^{2}

P = \frac{mgh}{t}

  = \frac{0.4*10*100}{20}

  = \frac{400}{20}

P = 20 W

3) u = 600 m/s

   g = 10 m/s^{2}

From the third equation of free fall,

V^{2} = U^{2} - 2gh

V is the final velocity, U is the initial velocity, h is the height.

0 = (600)^{2} - 2 x 10 x h

0 = 360000 - 20h

20h = 360000

h = \frac{360000}{20}

  = 18000

h = 18 km

The maximum height of the bullet would be 18 km.

3 0
3 years ago
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