Answer:
Coefficient of friction will be 0.587
Explanation:
We have given mass of the car m = 500 kg
Distance s = 18.25 m
Initial velocity of the car u = 14.5 m/sec
As the car finally stops so final velocity v = 0 m/sec
From second equation of motion
We know that acceleration is given by
So coefficient of friction will be 0.587
The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that
For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration
T= I α
0.5= 0.25 α
For Wheel B
m= 1 kg
d= 2 m,r=1 m
Given that angular acceleration is same for both the wheel
T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
Acceleration means any change in the speed or direction of motion.
