W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
The answer would be B: The ground exerts an equal force on the golf ball.
I’m not sure if its correct but I think it’s focal Ray point
For concave mirrors, some generalizations can be made to simplify ray construction. They are: An incident ray traveling parallel to the principal axis will reflect and pass through the focal point. An incident ray traveling through the focal point will reflect and travel parallel to the principal axis.
Because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) 2π12sqrt(9.81)=6.94922
