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patriot [66]
2 years ago
7

A 1400 kg car traveling in the positive direction takes 11.0 seconds to slow from 25.0 meters per second to 12.0 meters per seco

nd. What is the average force on the car during this time?
Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer:

–1652 N. The negative sign indicate that the force is in opposite direction to car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 1400 kg

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Force (F) applied on the car =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Acceleration (a) =?

a = (v – u) /t

a = (12 – 25) /11

a = –13/11

a = –1.18 m/s²

Finally, we shall determine the force on the car as follow:

Acceleration (a) = –1.18 m/s²

Mass (m) of car = 1400 kg

Force (F) applied on the car =?

F = ma

F = 1400 × –1.18

F = –1652 N

Thus, the force on the car is –1652 N. The negative sign indicate that the force is in opposite direction to car.

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A cheetah starts from rest and accelerates after a gazelle at a rate of 6.5 meters per second2for 3.0 seconds. Calculate the che
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the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s

Explanation:

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vf = vi + a * t

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5 0
3 years ago
An image of the moon is focused onto a screen using a converging lens of focal length f= 34.3 cm. The diameter of the moon is 3.
Rashid [163]

Answer:

The diameter of the moon's image is 0.31 cm.

Explanation:

Given that,

Focal length = 34.3 cm

Diameter of the moon d=3.48\times10^{6}\ m

Mean distance from the earth d=3.85\times10^{8}\ m

At that distance the object is assumes to be at infinity. hence the image will be formed at a distance equal to focal length

So, the image distance is 34.3 cm.

We need to calculate the  diameter of the moon's image

Using formula of magnification

m= \dfrac{image\ distance}{object\ distance}=\dfrac{height\ of\ image}{height\ of\ object}

\dfrac{0.343}{3.85\times10^{8}}=\dfrac{h'}{3.48\times10^{6}}

h'=\dfrac{0.343\times3.48\times10^{6}}{3.85\times10^{8}}

h'=0.0031\ m= 0.31\ cm

Hence, The diameter of the moon's image is 0.31 cm.

7 0
3 years ago
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