Answer:
The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².
How many times stronger than gravity is this force? 2.79 x 10⁴ g.
Explanation:
given information:
s = 220 m
final speed, vf = 10.97 km/s = 10970 m/s
g = 9.8 m/s²
he unrealistically large acceleration experienced by the space travelers during their launch
vf² = v₀²+2as, v₀ = 0
vf² = 2as
a =vf²/2s
= (10970)²/(2x220)
= 2.7 x 10⁵ m/s²
Compare your answer with the free-fall acceleration
a/g = 2.7 x 10⁵/9.8
a/g = 2.79 x 10⁴
a = 2.79 x 10⁴ g
Explanation:
It is known that relation between velocity and height is as follows.
v =
where, g = acceleration due to gravity = 9.8
h = height = 0.2 m
Therefore, velocity is calculated as follows.
v =
=
= 3.92 m/s
Also,
Putting the given values into the above formula as follows.
=
=
= m/s
or, = m/s
Thus, we can conclude that recoil speed of the Earth is m/s.
Answer:
500 N
Explanation:
Given that,
The upward force is 800 N and the downward forces are 400 N, 500 N, 400 N.
At equilibrium, the upward forces will become equal to the downward forces. Let the reading in the right hand scale.
x + 800 = 400 + 500 + 400
x + 800 = 1300
x = 1300 - 800
= 500 N
So, the reading in the right hand scale is 500 N.
I have all the answers here so take this
Answer:
3 m/s
Explanation:
First, find the time it takes to land.
Given in the y direction:
Δy = 0.8 m
v₀ = 0 m/s
a = 10 m/s²
Find: t
Δy = v₀ t + ½ at²
0.8 m = (0 m/s) t + ½ (10 m/s²) t²
t = 0.4 s
Next, find the velocity needed to travel 1.2 m in that time.
Given in the x direction:
Δx = 1.2 m
a = 0 m/s²
t = 0.4 s
Find: v₀
Δx = v₀ t + ½ at²
1.2 m = v₀ (0.4 s) + ½ (0 m/s²) (0.4 s)²
v₀ = 3 m/s