Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)

a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)

a = 7.67 ms⁻²
Answer:
A. 50 m/s
Explanation:
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 4 s
Find: v
v = at + v₀
v = (10 m/s²) (4 s) + 0 m/s
v = 40 m/s
In the x direction, the velocity is constant at 30 m/s.
The overall speed is:
v² = (30 m/s)² + (40 m/s)²
v = 50 m/s
Answer:
the answer to the question is the part that is reflected by clouds and atmosphere 26%
Answer:
The magnitude of angular acceleration is
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is
. Hence, this is the required solution.
Because even though the moon is smaller, therefore a weaker gravitational pull, the moon is much closer to the earth than the sun, thus having a greater gravitational pull