The speed of the sport car at the time of impact is 6.61 m/s.
<h3>What is the frictional force of the two cars?</h3>
Frictional force of the two cars = coefficient of kinetic frictin × mass × acceleration of gravity
= 0.8 × (2500+940) × 9.8
= 26970N
<h3>What is the acceleration of the skidded cars?</h3>
- As per Newton's second law of motion, force = mass × acceleration
- Acceleration= force / mass
= 26,970/3440
= 7.8 m/s²
<h3>What is the velocity of the sport car at the time of impact?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- Here, V = 0 m/s, a= -7.8 m/s², S= 2.8 m
- So, 0²-U²= 2×(-7.8)×2.8
=> U = √43.68
= 6.61 m/s
Thus, we can conclude that the speed of the sport car at the time of impact is 6.61 m/s.
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
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Answer:
did you mean released?
Explanation:
If so the process is called respiration
Newton 3rd law:
F=ma
1600=85*a
1600/85=a
a=18.82 m/s^2
Answer:
Power= 6.84×10⁸ W
Explanation:
Given Data
Niagara falls at rate of=1.4×10⁶ kg/s
falls=49.8 m
To find
Power Generated
Solution
Regarding this problem
GPE (gravitational potential energy) declines each second is given from that you will find much the kinetic energy of the falling water is increasing each second.
So power can be found by follow
Power= dE/dt = d/dt (mgh)
Power= gh dm/dt
Power= 1.4×10⁶ kg/s × 9.81 m/s² × 49.8 m
Power= 6.84×10⁸ W
Answer:
0.05 cm
Explanation:
The compression of the original spring = 12 - 8.55 cm = 3.45 cm = 0.0345 m
By Hooke's law, F = ke
Where F is the applied force, k is the spring constant and e is the extension or compression. In the question, F is the weight of the car.
k = F/e = 1355 × 9.8 / 0.0345 = 384898.55 N/m
This is the spring constant of the original spring. The question mentions that the force constant of the new spring is 5855.00 N/m smaller. Hence, the force constant of the new spring is 384898.55 - 5855 = 379043.55 N/m
With the new spring installed, the compression will be
e = F/k = 1355 × 9.8 / 379043.55 = 0.035 m = 3.5 cm
The difference in the compressions of both springs = 3.5 cm - 3.45 cm = 0.05 cm