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qaws [65]
3 years ago
5

Analyze why a body undergoing uniform motion eventually comes to rest even in the

Physics
1 answer:
djverab [1.8K]3 years ago
8 0

Answer:

It happens due to force of friction

Explanation:

If a body is performing a uniform motion and no external unbalanced force appears to apply on it, then  the body will come to rest after certain time. The reason behind this is the force of friction that is applied in opposite direction of the motion. So, when there is no apparent unbalanced force it means that the only force acting on the body is the force of friction. This force of friction tends to stop the motion after some period of time, because it is acting in the direction opposite to that of motion.

Hence, the reason behind a body undergoing uniform motion eventually stops is <u>Force of Friction.</u>

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doughnut-shaped chamber called the tokamak. This is where the fusion reactions take place, within hot plasma containing deuterium and tritium atoms.

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calculate the distance in km that alex can run in 2.0 hours if she maintains the constant average speed of 8.00 km/h
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A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch
Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0
3 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
1 year ago
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