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
I think its B B)Warm water rises and cold water moves in to replace it.
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
