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myrzilka [38]
3 years ago
14

If we approximate the rack to be completely flat and the racecar is travelling a constant 30.5 m/s around the turn, what forces

are responsible for the centripetal acceleration?
Physics
1 answer:
WITCHER [35]3 years ago
3 0

We have,

The race car is travelling a constant 30.5 m/s around the turn. When the car is travelling around a turn, the centripetal force acts on it. The centripetal force is balanced by the force of friction between the car and the surface. Hence, frictional force is responsible for the centripetal acceleration.

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Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h
Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

\Delta U = mg \Delta h

where

m is the man's mass

g is the gravitational acceleration

\Delta h is the change in height of the man

In this problem, we have:

\Delta U=4620 J is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

\Delta h=8.4 m is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg

6 0
3 years ago
Read 2 more answers
A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v
slava [35]

<u>Answer:</u>

 The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

<u>Explanation:</u>

  Initial location of spaceship = (600 i - 400 j + 200 k)*10^3km= (600 i - 400 j + 200 k)*10^6m

  Initial velocity = 9500 i m/s

  Acceleration = (40 i - 20 k)10^3m/s^2

  Time = 35 minute = 35 * 60 = 2100 seconds

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  Substituting

       s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m

     So final position = ((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m

                              =(708.15 i - 444.1 j + 200 k)*10^3km

    The spaceship's position when the engine shuts off = (708.15 i - 444.1 j + 200 k)*10^3km

3 0
3 years ago
Read 2 more answers
Identify the characteristics of protists. Check all that apply.
romanna [79]

Answer:

There are no examples

Explanation:

3 1
3 years ago
Read 2 more answers
What are the examples of non viscous liquid​
liq [111]
<h2><u>ANSWER</u></h2>

\\

Two isotopes of helium ( helium 3 and helium 4) show superfluidity whem they are liquified by cooling to a very low specific temperature. One isotope of rubidium and lithium posses this property at low temperature.

Hope It Helps!

8 0
3 years ago
Due to the doppler effect when moving away from the source of a sound
Ray Of Light [21]
You do not hear it as clearly as you did, the frequency goes down. Likewise, if you moved closer you would hear it more.
5 0
3 years ago
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