Answer:
E = 2.7 x 10¹⁶ J
Explanation:
The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:
where,
E = Energy Released = ?
m = mass of material reduced = 0.3 kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>E = 2.7 x 10¹⁶ J</u>
There isnt enough information to answer the question, the missing variable is "distance from said falling spot and ground"
Answer:
a.) L = 2.64 kgm^2/s
b.) V = 4.4 m/s
Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.
So,
Radius r = 0.6 m
Mass M = 2 kg
Velocity V = 1.1 m/s
Angular momentum L can be expressed as;
L = MVr
Substitute all the parameters into the formula
L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1
the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1
b. If she pulls her arms into 0.15 m,
New radius = 0.15 m
Using the same formula again
L = 2( MVr)
2.64 = 2( 2 × V × 0.15 )
1.32 = 0.3 V
V = 1.32/0.3
V = 4.4 m/s
Her new linear speed will be 4.4 m/s
