Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as
Where 
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and = 
= 9.231 N
Therefore the acceleration due to gravity = /mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
5 times that of initial pressure i.e 1625 kpa
<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:
F = G(Mm / (4(pi)*r^2))
The above expression gives the force that you feel on the earth's surface, as it is today!
Let us now double the mass of the earth and decrease its diameter to half its original size.
This is the same as replacing M with 2M and r with r/2.
Now the gravitational force (F' ) on the new earth's surface is given by:
F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F
So:
F' = 8F
This implies that the force that you would feel pulling you down (your weight) would increase by 800%!
You would be 8 times heavier on this "new" earth!</span>
