A vertical ideal spring is mounted on the floor and has a spring constant of 116 N/m. A 0.60-kg block is placed on the spring in
1 answer:
Answer:
a
b
Explanation:
From the question we are told that
The spring constant is
The mass of the block is
Considering question a
Generally the weight of the block is mathematically represented as
=>
=>
Generally the force exerted on the spring is mathematically represented as
At equilibrium
=>
=>
=>
Considering question b
Generally the energy stored in the spring is mathematically represented as
Now the potential energy of the block before it is drooped is mathematically represented as
Generally from the law of energy conservation we have that
=>
=>
=>
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Answer:
(A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
Explanation:
Given that,
Distance = 20.0 m
Frictional force = 35.0 N
Angle = 25.0°
(A). We need to calculate the work done on the cart by the friction
Using formula of work done
Where, F = force
d = distance
Put the value into the formula
(B). The work done by the gravity is perpendicular to the direction of the motion
We need to calculate the work done on the cart by the gravitational force
Using formula of work done
Put the value into the formula
(C). We need to calculate the work done on the cart by the shopper
Using formula of work done
Put the value into the formula
(D). We need to calculate the force the shopper exerts
Using formula of force
Put the value into the formula
(E). We need to calculate the total work done on the cart
Using formula of work done
Put the value into the formula
Hence, (A). The work done on the cart by the friction is -700 J.
(B). The work done on the cart by the gravitational force is 0 J.
(C). The work done on the cart by the shopper is 700 J.
(D). The force the shopper exerts 38.61 N.
(E). The total work done on the cart is zero.
Answer:
Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.
Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law
where are the refractive index of the 1st and 2nd medium
are the angle that the incident ray and the refracted ray makes with the normal to the interface
In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).
In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (), this means that
, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (
), so it bends away from the normal (
).
In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h
However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).
Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.
According to the law of reflection:
- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane
- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)
Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.