1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m
Answer:
Explanation:
Given
Displacement is 
of Amplitude
i.e. 
, where A is maximum amplitude
Potential Energy is given by
Total Energy of SHM is given by
Total Energy=kinetic Energy+Potential Energy
Potential Energy is 
th of Total Energy
Kinetic Energy is 
of Total Energy
(c)Kinetic Energy is 
Answer:
2.24 m/s
Explanation:
resolving force of 29.2 N in x component
Fx = 29.2 cos 57.7
Fx = 15.6N
as force of friction is 12.7 N hence net force which produces acceleration is
15.6-12.7=2.9 N
by Newton 's law a=f/m
a= 2.9/6.87=0.422 m/s^2
now equation of motion is
v^2= U^2+2as
= 0^2+2(.422)(5.93)
v^2=5.00
v=2.24 m/s
Yeah sure someone else in answering rn
