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Lera25 [3.4K]
3 years ago
9

When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of

static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Physics
1 answer:
pentagon [3]3 years ago
7 0

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

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Answer:

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Explanation:

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4 0
3 years ago
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Why is it that an object can accelerate while
posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

And any change of velocity is called acceleration.

3 0
3 years ago
How can you increase the force required to move an object without changing the mass of the object?
Effectus [21]

To increase the force required to move an object without change mass is : To increase the acceleration of the object

<h3>Force acting on an object </h3>

Given that Force = Mass * acceleration

An increase in the acceleration of an object in motion will result in a proportional increase in the Force required to move the object becasue force of an object is directly proportional to the acceleration of an object.

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8 0
2 years ago
Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe
yulyashka [42]

Explanation:

Mass of two soccer balls, m_1=m_2=0.4\ kg

Initial speed of first ball, u_1=0

Initial speed of second ball, u_2=3.5\ m/s

After the collision,

Final speed of the second ball, v_2=0

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v_1 is the final speed of the first ball

0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0

0.4\times 3.5=0.4v_1

v_1=3.5\ m/s

(b) Let E_1 is the kinetic energy of the first ball before the collision. It is given by :

E_1=\dfrac{1}{2}mu_1^2

E_1=\dfrac{1}{2}\times 0.4\times 0

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.

7 0
3 years ago
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Help help me please!!!!!
Zanzabum

Answer:

I think that its a or b

Explanation:

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