Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
<u />
Answer:
It is (1/5)th as much.
Explanation:
If we apply the equation
F = G*m*M / r²
where
m = mass of a man
M₀ = mass of the planet Driff
M = mass of the Earth
r₀ = radius of the planet Driff
r = radius of the Earth
G = The gravitational constant
F = The gravitational force on the Earth
F₀ = The gravitational force on the planet Driff
g = the gravitational acceleration on the surface of the earth
g₀ = the gravitational acceleration on the surface of the planet Driff
we have
F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²
⇒ F₀ = G*m*M / (5*r²) = (1/5)*F
If
F₀ = (1/5)*F
then
W₀ = (1/5)*W ⇒ m*g₀ = (1/5)*m*g ⇒ g₀ = (1/5)*g
It is (1/5)th as much.
15 miles to kilometers would be: 24.14 kilometers