We want to study the impact of a sledgehammer and a wall.
Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.
When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.
This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.
Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.
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Explanation:
I will do two of each as examples.
Boyle's law says that at constant temperature, the product of the initial pressure and volume equals the product of the final pressure and volume.
1. P₁ V₁ = P₂ V₂
(1.5 atm) (10.0 L) = (0.75 atm) V
V = 20.0 L
2. P₁ V₁ = P₂ V₂
(100.0 kPa) (500.0 mL) = P (1,000.0 mL)
P = 50.0 kPa
Charles' law says that at constant pressure, the quotient of the initial volume and temperature equals the quotient of the final volume and temperature.
6. V₁ / T₁ = V₂ / T₂
(10.0 L) / (1500 K) = V / (750 K)
V = 5.0 L
7. V₁ / T₁ = V₂ / T₂
(500.0 mL) / (100 K) = (1000.0 mL) / T
T = 200 K
Answer:primary current is 3A
Explanation:
primary voltage Vp=240V
secondary voltage Vs=60V
resistance 5Ω
primary current in the circuit is
K.E =1/2mv2
M=6kg
V=3m/s
K.E=1/2 X 6 X 3 X 3
=1/2 X 6 X 9
=27 J
Answer:
W = 30.38 N
Explanation:
Given that,
Mass of a rock, m = 3.1 kg
We need to find the weight of the rock on the surface of Earth. Weight of an object is given by :
W = mg
g is the acceleration due to gravity, g = 9.8 m/s²
W = 3.1 kg × 9.8 m/s²
= 30.38 N
So, the weight of the rock on the Earth is 30.38 N.
