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.
Carbon: C, 12.011, 6, 12
Oxygen: O, 8, 8, 8, 16
Boron: B, 10.811, 5, 5, 11
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
The distance between the two charges is 
Explanation:
The electrostatic force between two charged objects is given by Coulomb's law:
where:
is the Coulomb's constant
are the charges of the two objects
r is the separation between the two charges
In this problem, we are given the following:
Therefore, we can rearrange the equation to solve for r, the distance between the two charges:
Learn more about electrostatic force:
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