Answer:
410 m
Explanation:
Given:
v₀ = 20.5 m/s
a = 0 m/s²
t = 20 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²
Δx = 410 m
Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
Answer:
Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
