Sure.
Can I use your answer to part-'a' ?
If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of
(32 rev/min²) x (1.2 min) = 38.4 rev/min .
Check:
If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is
(1/2) (0 + 38.4) = 19.2 rpm average
At an average speed of 19.2 rpm for 1.2 min,
it covers
(19.2 rev/min) x (1.2 min) = 23.04 revs .
That's pretty close to the "23" in the question, so I think that
everything here is in order.
Gravitational potential energy can be described as m*g*h (mass times gravity times height).
Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.
After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.
To find how much energy was added, we subtract final energy from initial energy:
147 J - 44.1 J = 102.9 Joules.
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
Answer:
The value is 
Explanation:
From the question we are told that
The initial speed is 
Generally the total energy possessed by the space probe when on earth is mathematically represented as
Here 
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=> 
=> 
And 
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
Here 
is the escape velocity from earth which has a value 
=> 
=> 
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
Generally from the law energy conservation we have that
So
=> 
=> 
=>
