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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Answer:
no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.
Answer:
The distance between the camera and the rock is 836.6 cm
Explanation:
A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:
cos α = adjacent/hypotenuse
cos(76.31) = 198.0/h
h = 198.0/cos(76.31)
h = 836.6 cm
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) / m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.
Answer:
3.2N
Explanation:
Given parameters:
Mass of block = 1.5kg
Coefficient of kinetic friction = 0.6
Force of pull on block = 12N
Unknown:
Net force on the block = ?
Solution:
Frictional force is a force that opposes motion:
Net force = Force of pull - Frictional force
Frictional force = umg
u is coefficient of kinetic friction
m is the mass
g is the acceleration due to gravity
Frictional force = 0.6 x 1.5 x 9.8 = 8.8N
Net force = 12N - 8.8N = 3.2N