Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Explanation:
It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².
The second equation of kinematics gives the relationship between the height reached and time taken by it.
Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.
We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :
t is time taken by the ball to hit the ground
is initial speed of the ball
So, the correct option is (A).
Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.
40 mi/min
