A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to t
1 answer:
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
You might be interested in
Answer:
Explanation:
Given
inclination
initial speed
Point of release is 45 m above the ground
Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is
Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30
where,
thus total time time required is
vertical velocity just before hitting
Horizontal velocity
Net velocity Just before hitting
Answer: The length of the shadow on the wall is decreasing by 0.6m/s
Explanation:
the specified moment in the problem, the man is standing at point D with his head at point E.
At that moment, his shadow on the wall is y=BC.
The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:
ADAB=DEBC
8/12=2/y,∴y=3 meters
If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.
(12−x) /12=2/y
1− (1 /12x )=2 × 1/y
Let's take derivatives of both sides:
−1 / 12dx = −2 × 1 / y^2 dy
Let's divide both sides by dt:
−1/12⋅dx/dt=−2/y^2⋅dy/dt
At the specified moment:
dxdt=1.6 m/s
y=3
Let's plug them in:
−1/121.6) = - 2/9 × dy/dt
dy/dt = 1.6/12 ÷ 2/9
dy/dt = 1.6/12 × 9/2
dy/dt = 14.4/24 = 0.6m/s
