Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J
The amount of current passing through the point is 1 A
The amount of current passing through the point can be calculated using the formula below.
⇒ Formula:
- Q = i/t......................... Equation 1
⇒ Where:
- Q = Charge
- i = current
- t = time.
⇒ Make "i" the subject of the equation.
- i = Qt....................... Equation 2
From the question,
⇒ Given:
- Q = 1000 millicoulombs = 1 coulombs
- t = 1 seconds. (Assuming the time is 1 seconds)
⇒ Substitute these values into equation 2
Hence, The amount of current passing through the point is 1 A.
Learn more about charges here: brainly.com/question/4158552
Answer:
50 N/m
Explanation:
Elastic energy = kinetic energy
EE = KE
½ kx² = ½ mv²
½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²
k = 50 N/m
Hypothesis: Enriched environments will have a positive effect and better performance of the experimental group of animals exposed to toys in comparison to the control group of animals in running in mazes.
Dependent Variable: Control Group
Independent Variable: Exposed/Experimental Group
Answer:
The potential energy stored in the spring is 0.018 J.
Explanation:
Given;
spring constant, k = 90 N/m
extension of the spring, x = 2 cm = 0.02 m
The potential energy stored in the spring is calculated as;
U = ¹/₂kx²
where;
U is the potential energy stored in the spring
Substitute the given values in the equation above;
U = ¹/₂ x 90 N/m x (0.02 m)²
U = 0.018 J
Therefore, the potential energy stored in the spring is 0.018 J.
