Subduction occurs in both locations
Answer:
a) Weight of the rock out of the water = 16.37 N
b) Buoyancy force = 4.61 N
c) Mass of the water displaced = 0.47 kg
d) Weight of rock under water = 11.76 N
Explanation:
a) Mass of the rock out of the water = Volume x Density
Volume = 470 cm³
Density = 3.55 g/cm³
Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg
Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N
b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.
Volume = 470 cm³
Density of liquid = 1 g/cm³
c) Mass of the water displaced = Volume of body x Density of liquid
Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg
d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force
Weight of rock under water = 16.37 - 4.61 =11.76 N
Answer:
9.6m/s
Explanation:
Using the equation S=d/t where s=speed, d=distance, and t=time
plug in the known variables
S=120m/12.5s
S=9.6m/s
The chemical energy in Jay's body, to kinetic energy in the car
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).
And you know that the potential energy, PE, is [ 1/2 ] k (x^2)
Then, use x = A, to calculate the PE in the point where ME = PE.
ME = PE = [1/2] k (A)^2.
At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2
=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME
So, if PE is 1/4 of ME, KE is 3/4 of ME.
And the answer is 3/4