Answer:
distance difference would a) increase
speed difference would f) stay the same
Explanation:
Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.
Their equations of motion for distance and velocities are




Their difference in distance are therefore:


(As

So as time progress t increases, Δs would also increases, their distance becomes wider with time.
Similarly for their velocity difference


Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.
This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;



W =
-cosθ ![]^{0.047}_{0.045 }](https://tex.z-dn.net/?f=%5D%5E%7B0.047%7D_%7B0.045%20%7D)
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
It would either break or stop depends on the density
Answer:
t₁ > t₂
Explanation:
A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁
Solution:
Newton's law of motion is given by:
s = ut + (1/2)gt²;
where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.
u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.
hence:
When stationary:

Hence t₂ < t₁, this means that t₁ > t₂.