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____ [38]
4 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]4 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 ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit
vichka [17]

As the ball is moving in air as well as we have to neglect the friction force on it

So we can say that ball is having only one force on it that is gravitational force

So the force on the ball must have to be represented by gravitational force and that must be vertically downwards

So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

<em>Diagram A shows a box with a downward arrow. </em>

8 0
3 years ago
A traveler pulls on a suitcase strap at an angle 36 above the horizontal with a force of friction of 8 N with the floor. If 752
goldenfox [79]

Answer:

71.8 N

Explanation:

T = Tension force in the strap

W = net work done = 752 J

f = force of friction = 8 N

d = displacement = 15 m

θ = angle between tension force and horizontal displacement = 36 deg

work done by frictional force is given as

W' = - f d

Work done by the tension force is given as

W'' = T d Cos36

Net work done is given as

W = W' + W''

W = T d Cos36 - f d

752 = T (15) Cos36 - (8) (15)

T = 71.8 N

8 0
3 years ago
Al is floating freely in her spacecraft, and you are accelerating away from her with an acceleration of 1g. 5) How will you feel
wariber [46]

Answer:

D. You will feel the same weight as you do on Earth

Explanation:

In free space, she is suppose to be weightless.

Free fall can be described as body in motion where the body is under the effect of acceleration due to gravity only and no other acceleration..

Since I am accelerating away from her at an acceleration of 1g

Then,

F=ma, where a=g

Then F=mg

Since my weight on earth is W=mg

This is equals to my weight in the spaceship, then I will feel the same weights as I do on earth.

5 0
3 years ago
Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se
ANTONII [103]

Answer:

Density

Explanation:

Each of the three liquids have different densities. Have you ever tried to mix oil and water? They don't mix because they have different densities.  

3 0
3 years ago
Read 2 more answers
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi
k0ka [10]

Answer:

a) F₁₂₀ = 1.34 pa A  , b)  F₂₀ = 0.746 pa A

Explanation:

Part. A .    The definition of pressure is

         P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

        P V = n R T

Let's write this equation for two temperatures

       P₁ V = n R T₁

       P₂2 V = n R T₂

       P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 393/293

       P₂ = 1.34 pa

We calculate the strength

       P₂ = F₁₂₀ / A

       F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

        P₁ / P₂ = T₁ / T₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 293/393

       P₂ = 0.746 pa

Let's calculate the force (F20), from this point

      F₂₀ / A = 0.746 pa

     F₂₀ = 0.746 pa A

6 0
3 years ago
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