The change in Potential energy of the cat is 176.4 J.
<h3 /><h3>Potential Energy:</h3>
This is the energy due to the position of a body. The S.I unit is Joules (J)
The formula for change in potential energy.
<h3 /><h3>Formula:</h3>
- ΔP.E = mg(H-h).............. Equation 1
<h3>Where:</h3>
- ΔP.E = Change in potential energy
- m = mass of the cat
- g = acceleration due to gravity
- H = First height
- h = second height.
From the question,
<h3>Given:</h3>
- m = 15 kg
- H = 2.5 m
- h = 1.3 m
- g = 9.8 m/s²
Substitute these values into equation 1
- ΔP.E = 15×9.8(2.5-1.3)
- ΔP.E = 15×9.8×1.2
- ΔP.E = 176.4 J.
Hence, The change in Potential energy of the cat is 176.4 J
Learn more about Potential energy here: brainly.com/question/1242059
I found this using the app Socratic. When I took physics in high school it helped me so much.
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
Answer: 2. 2.0*10^2 W
Explanation:
Power = Work/Time
Power = (2.0*10^3) Joules/10 seconds
Power = 2.0*10^2 Watts
Answer:
<em>Its speed will be 280 m/s</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.
If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

The object accelerates from rest (vo=0) at a constant acceleration of
. The final speed at t=35 seconds is:


Its speed will be 280 m/s