Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
Answer: The tidal forces exerted by the moon are directly associated with the earth's rotation. Due to the strong gravitational pull of the moon, the tidal bulging appears on both the sides on earth and these are region of high tide, and there is gradual rise and fall of sea level.
Because of these tidal effect, the earth is able to rotate only once in each of the orbital period.
