Answer:c
Explanation:
Given
object is falling Freely with an odometer
Suppose it falls with zero initial velocity
so distance fallen in time t is given by

here u=0 and t=time taken

for 

for 

distance traveled in 2 nd sec
for 

distance traveled in 3 rd sec
so we can see that distance traveled in each successive second is increasing
Answer:
Second option 6.3 N at 162° counterclockwise from
F1->
Explanation:
Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.
For the address x we have:

For the address and we have:

The forces
and
are known

We have 2 unknowns (
and b) and we have 2 equations.
Now we clear
from the second equation and introduce it into the first equation.

Then

Then we find the value of 

Finally the answer is 6.3 N at 162° counterclockwise from
F1->
Answer:
Fighting
Explanation:
Fighting with someone like punch for punch can make the person bleed which an injury to the guy who you fought with and you can be arrested if the parents file a complain at the police station
Answer:
Also 3s.
Explanation:
Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, 
We need to find the thrust and the force. The mass of the liquid displaced is given by :

V is volume
Weight of the displaced liquid
W = mg

So,

So, the thrust and the force is 1562.53 N.