Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
Answer:
Explanation:
We can solve the problem by using the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the car in this problem:
u = 0 (it starts from rest)
is the final velocity
s = 10 km = 10 000 m is the displacement
Solving for a, we find:
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Answer:
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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
