1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kirza4 [7]
3 years ago
12

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of

curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?
Physics
1 answer:
Alik [6]3 years ago
6 0

This question is incomplete, the complete question is;

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.

(a) What is the force (N) that the track must exert on the car? (positive is up)

(b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answer:

a) the force (N) that the track must exert on the car is -6139.14 N

b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

Explanation:

Given the data in the question;

Let N represent the force that the track must exerted on the car

Net force on the car Fnet = Mg + N

so

M × a = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

we substitute

N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )

N = 3136 - ( 320 × 28.9848 )

N = 3136 - 9275.136

N = -6139.14 N

Therefore, the force (N) that the track must exert on the car is -6139.14 N

b) What must be the force (N) that the car exerts on a 61 kg passenger?

Let N represent the force that the car exerts on 61kg passengers

so

Net force of passengers Fnet = mg + N

Ma = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )

N = 597.8 - ( 61 × 28.9848)

N = 597.8 - 1768.0728

N = -1170.27 N

Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

You might be interested in
A force of 600 N is acting on a motorcycle that has a mass of 240 kg. What is the acceleration of the motorcycle?
Verdich [7]

Answer:

2.5m/s2

Explanation:

The following were obtained from the question:

F = 600N

M = 240 kg

a =?

Recall: F = Ma

a = F/M

a = 600/240

a = 2.5m/s2

Therefore, the acceleration of the motorcycle is 2.5m/s2

7 0
3 years ago
Read 2 more answers
If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo
Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0
2 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
3 years ago
Which is an example of kinetic energy? A. a stretched rubber band B. wind C. water in a reservoir D. natural gas E. an object su
daser333 [38]

Kinetic energy is energy of motion.

In the cases of a stretched rubber band, water in a reservoir, natural gas, or an object suspended above the ground, everything is just laying there, and nothing is moving. There's nothing there that has kinetic energy.

If there's any wind, then air is moving. The moving air has kinetic energy.

6 0
3 years ago
How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?
allochka39001 [22]

b 250 j

hope  this helps                            

6 0
3 years ago
Read 2 more answers
Other questions:
  • While punting a football, a kicker rotates his leg about the hip joint. the moment of inertia of the leg is 3.75 kg m2 and its r
    13·2 answers
  • Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer
    15·1 answer
  • How to find the frictional force acting on an object (not the friction coefficient)? ...?
    6·1 answer
  • In the illustration, which two simple machines are being used to enable the student to reach the door?
    12·2 answers
  • What is the weight of a 75kg. Mass?
    8·2 answers
  • A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
    10·1 answer
  • A fisherman sees 7 wave crests go by in 10.0 s. The crests are 2.43 m apart. Find the period of the wave. (Unit = s)​
    11·2 answers
  • 1) Use SolidWorks (SW) FEA to apply a bending load of 600 lbf on the right end of the stepped shaft as shown below. This is the
    9·1 answer
  • I really need some with help with this anyone who can help with acceleration and velocity pleasee and thank you
    8·1 answer
  • A=100m/s2<br> Vi=10m/s<br> T=10sec<br> Vf?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!