Answer:
The gravitational potential energy of that rock is 174371.4 J.
Explanation:
Given
To determine
We need to find the gravitational potential energy of the rock
We know that the potential energy of a body is termed as the stored energy due to its position.
One kind of energy comes from Earth's gravity — Gravitational potential energy (GPE).
Gravitational potential energy (GPE) can be determined using the formula
where
is the mass
is the gravitational acceleration which is equal to g = 9.8 m/s²
is the height
- GPE is the Gravitational potential energy
now substituting m = 59.31, h = 300 and g = 9.8
J
Therefore, the gravitational potential energy of that rock is 174371.4 J.
Answer:
B) 2.7 g of aluminium has a volume of 1 cm^3
Explanation:
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³
Check:
Given the following data;
Mass = 2.7 grams
Volume = 1 cm³
Substituting into the formula, we have;
Density = 2.7 g/cm³
Answer:
-2.5m/s²
Explanation:
The acceleration of a body is giving by the rate of change of the body's velocity. It is given by
a = Δv / t ----------------(i)
Where;
a = acceleration (measured in m/s²)
Δv = change in velocity = final velocity - initial velocity (measure in m/s)
t = time taken for the change (measured in seconds(s))
From the question;
i. initial velocity = 5m/s
final velocity = 0 [since the body (ball) comes to rest]
Δv = 0 - 5 = -5m/s
ii. time taken = t = 2s
<em>Substitute these values into equation (i) as follows;</em>
a = (-5m/s) / (2s)
a = -2.5m/s²
Therefore, the acceleration of the ball is -2.5m/s²
NB: The negative sign shows that the ball was actually decelerating.
Answer: B
Explanation:the voltage is just like the force that drives the current through out the circui... When trippled, the force increases and the current increases since the resistance in the circuit remains constant.
