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seropon [69]
3 years ago
15

A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the

wheels and the track is 0.19. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track? Answer in units of m/s.
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

coefficient of static friction, μ = 0.19

acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

v_{max}=\sqrt{\mu Rg}

v_{max}=\sqrt{0.19\times53.3\times9.8

vmax = 9.96 m/s

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A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what
Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

P=m*v\\

where:

m = mass [kg]

v = velocity [m/s]

P =20*10\\P =200 [kg*m/s]

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

v = P/m\\v = 200/5\\v = 40 [m/s]

3 0
3 years ago
If a body starts moving from rest its<br>initial velocity is<br>ų <br>V<br>zero<br>m/s​
Artist 52 [7]

Answer:

zero

When a Body start a from rest ,its initial velocity is zero.

6 0
3 years ago
Object a and object b are both in motion when they collide with each other. They then continue in a new direction unaffected by
Harrizon [31]
This is an elastic collision

bcuz i think they move apart after the collision

sorry if im wrong
8 0
3 years ago
A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t
Elodia [21]

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

5 0
3 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
3 years ago
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