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____ [38]
3 years ago
7

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

al force required to keep the faster car on the road without skidding is F, then the frictional force required to keep the slower car on the road without skidding is __-
Physics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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A 250 g beach ball rolls across the sand with a speed of 11.16 km/h. First convert units to kg and m/ then determine the momentu
yuradex [85]

Answer:

0.775 kg-m/s

Explanation:

Convert the units to the right unit forms necessary

250 g -> 0.25 kg

11.16 km/h -> 3.1 m/s

Now use the formula:

                         velocity

                   mass /  

momentum  /    /

       \           /    /

         \       /    /

          p = mv

p = 0.25 × 3.1 = 0.775 kg-m/s

Hope this helps you!

Bye!

6 0
3 years ago
A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.
Bond [772]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

4 0
3 years ago
What is it called when something doesn’t flow easy
LUCKY_DIMON [66]

Answer:

That something is called having a high viscosity.

Explanation:

The measure expressing a "resistance to flow" is called viscosity. Viscosity relates to  internal friction forces in a fluid causing it to flow with more or less difficulty. Highly viscous stuff is perceived "thick" or "sticky."

7 0
3 years ago
The force exerted on the tires of a car that directly accelerate it along a road is exerted by the
azamat

The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

<h3>What is force?</h3>

Force is defined as the product of mass and acceleration of an object.

Friction is defined as the force that resists the movement of an object over another.

Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

Learn more about force here:

brainly.com/question/12970081

#SPJ12

7 0
2 years ago
A 2.6 kg mass attached to a light string rotates on a horizontal,
Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

  • <em>maximum mass the string can support before breaking, m = 17.9 kg</em>
  • <em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0
3 years ago
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