Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²
There is one mistake in the question.The Correct question is here
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Answer:

Explanation:
From the question we are told that:
Bowling ball Speed 
Beach ball Speed 
Let The Mass be equal i.e

Therefore
Generally the equation for Velocity of beach ball after collision
is mathematically given by
Since Velocity is Vector Quantity
Therefore


