M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).
The momentum before colision is equal with the momentum after colision:
m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem
=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
To be effective, an exercise program must have an aerobic
form, portion for strength enhancement, and a stretching part. These three
things are essential because they each target specific improvements in your
body. For example, aerobics can help you maintain your body’s fitness or make
it better. This usually targets your heart rate and ensures that you burn fat
while doing so. Second is strength enhancement; this will make sure that your
body becomes better – not just in a feeble state. Lastly is stretching, your
muscles are like rubber bands. You cannot end or start your exercise program
without stretching simply because they can damage your muscles as well. Aside
from this, stretching can stop you from shocking your body into a physical
activity, which may cause you to lose consciousness or have undue stress and fatigue.
Answer:
the period of the 16 m pendulum is twice the period of the 4 m pendulum
Explanation:
Recall that the period (T) of a pendulum of length (L) is defined as:
where "g" is the local acceleration of gravity.
SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:
therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.
Answer:
=24.25 ^−1
Explanation:
Let and be initial and final velocity of the body respectively,
be acceleration due to gravity ( 9.8^−2 ), ℎ be the height of the body.
=0 ^ −1
ℎ=30
we know that, ^2−^ 2=2ℎ
^2=2∗9.8∗30
^2=588
=24.25 ^−1
Answer: 
Explanation:
Given
Wavelength of light 
Screen is 
away
Distance between two adjacent bright fringe is 
When same experiment done in water, wavelength reduce to 
So, the distance between the two adjacent bright fringe is 
Keeping other factor same, distance becomes
