Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.
Answer:
Acceleration of the bullet will be 1778835.6
Explanation:
We have given length of the barrel refile s= 0.855 m
When the bullet leaves the muzzle its velocity is 553 m/sec
So final velocity v = 553 m/sec
Initial velocity will be 0 that is u = 0 m/sec
According to third equation of motion
Answer
given,
SAT is 500 with a standard deviation of 100.
a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.
sample mean =
=
= 5
95% confidence level is achieved within +/- 1.960 standard deviations.
1.960 standard deviations x 5 is equal to +/- 9.8
confidence interval = 511 - 9.8 --- 511 + 9.8
= 501.2-----520.8
Here the weight is hung from the thread inside the ceiling of train
now the FBD of train is analysed
We can say
now by ratio of two equations we have
this shows that train must be in accelerating condition due to which the thread is making an angle with the vertical
now we will have
2 The train is not an inertial frame of reference.
Since train is accelerating so it must be non inertial frame
5 The train may be moving at a constant speed in a circle.
Since its moving in circle so it is in accelerating condition
8 The train must be accelerating.