The net force will point towards the acceleration of the object, as supported by Newton's second law.
The formula for the mass that remains:
m₀ - the initial mass, t - time, T - the half-life
The answer is c. 1.25 g.
Answer: -3.49 m/s (to the south)
Explanation:
This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum 
must be equal to the final momentum 
, and taking into account this is aninelastic collision:
Before the collision:
(1)
After the collision:
(2)
Where:
is the mass of the car
is the velocity of the car, directed to the north
is the mass of the truck
is the velocity of the truck, directed to the south
is the final velocity of both the car and the truck
(3)
(4)
Isolating :
(5)
(6)
Finally:
The negative sign indicates the direction of the velocity is to the south
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:
For every action, there is an equal and opposite reaction.
Explanation:
Physics helps alot lol
