Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
Answer:
inversion would be going toward or inside internal and eversion is going outside or external
Answer:
In an elastic collision, the total kinetic energy is conserved, while in an inelastic collision, it is not
Explanation:
Let's define the two types of collision:
- Elastic collision: an elastic collision is a collision in which:
1) the total momentum of the system is conserved
2) the total kinetic energy of the system is conserved
Typically, elastic collisions occur when there are no frictional forces acting on the objects in the system, so that no kinetic energy is lost into thermal energy. An example of elastic collision is the collision between biliard balls.
- Inelastic collision: an inelastic collision is a collision in which:
1 ) the total momentum of the system is conserved
2) the total kinetic energy of the system is NOT conserved
In an elastic collision, part of the total kinetic energy is lost (=converted into thermal energy) due to the presence of frictional forces. An example of inelastic collision is the accident between two cars, in which part of the energy is converted into heat.
Answer:
The weight of the block on the moon is 15 kg.
Explanation:
It is given that,
The acceleration of the block, a = 7.5 m/s²
Force applied to the box, F = 70 N
The mass of the block will be,
m = 9.34 kg
The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :
W = 15.13 N
or
W = 15 N
So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.
Answer: x ≈ 36.3 cm
Explanation:
Conservation of momentum during the collision
0.0340(120) + 1.24(0) = (0.0340 + 1.24) v
v = 3.2025 m/s
The kinetic energy of the block/bullet mass will convert to spring potential
½kx² = ½mv²
x = √(mv²/k)
x = √(1.274(3.2025²) / 99.0)
x = 0.363293... ≈ 36.3 cm