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horsena [70]
4 years ago
10

A car accelerates at 3 m/s*2. Assuming the car starts from rest, how much time does it need to

Physics
2 answers:
Eduardwww [97]4 years ago
6 0

We are given with some parameters used in Kinematics for determining the motion of an object.

  • Accleration = 3 m/s²
  • Started from rest (Initial vel.) = 0 m/s
  • Final velocity = 12 m/s

We have to find the time taken by the car to attain the final velocity with the given accleration. By using 1st equation of motion which says:

  • v = u + at

Final velocity = Initial velocity + Accl. × Time taken

Plugging the values,

➝ 12 = 0 + 3(t)

➝ 3t = 12

➝ t = 4 s

So, the correct answer is Option C. And we are done...

Carry On Learning !

uranmaximum [27]4 years ago
5 0
<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the time take by a car.

<h3><u>Solution:-</u><u> </u></h3>

According to the first equation of motion:

v = u + at

★ Substituting the values in the above formula,we get:

⇒ 12 = 0 + 3 × t

⇒ 12 = 3t

⇒ 3t = 12

⇒ t = 12/3

⇒ t = 4 sec

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3 years ago
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(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface
Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

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Acceleration due to gravity is given by

g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}

The expression of escape velocity is given by

v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s

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3 years ago
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A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t
levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

F=m\omega^2 r

where

m is the mass of the object

\omega is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s is the angular velocity

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Substituting, we find the force:

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shtirl [24]

Answer:

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