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mariarad [96]
3 years ago
10

a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h

as a radius of 80 meters. What is the centripetal force on the car
Physics
2 answers:
DochEvi [55]3 years ago
6 0

your answer is 15,625 N

i just took the test

melisa1 [442]3 years ago
4 0
In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
                            = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 1250000/80
                            = 15625
So the centripetal force on the car is 15625 Newtons
  
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What is an electric circuit?
agasfer [191]

Answer:

the answer to the question is a system of insulating element designed to control the path of electric current for a particular purpose

5 0
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If there was no frictional force acting on the ball in question 2 what would happen to the player? Use inertia and Newton’s 1st
laila [671]
Newwton's law of inertia states that an object will not be able to move unless force is applied to it
6 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 47 N. How f
Novosadov [1.4K]

Answer:

7.53 m

Explanation:

Force, F = 47 N

initial velocity, u = 0

Final kinetic energy, Kf = 354 J

Let the distance traveled by the student is s.

According to the work energy theorem,

Work done by all forces = Change in kinetic energy

Force x distance = final kinetic energy - initial kinetic energy

F x s = kf - ki

47 x s = 354

s = 7.53 m

4 0
3 years ago
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