Answer:
The distance of separation is decreased
Explanation:
From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
Answer:
Explanation:
The work done is defined as the product of force applied in the direction of displacement and the displacement.
W = F x d x Cosθ
where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.
For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.
Answer:
Weight of the car, normal force, drag force
Explanation:
The forces acting on the car are:
- The normal force which acts perpendicularly to the downhill plane
- The weight of the car which acts vertically downwards
- The drag force due to air resistance which acts in opposition to the motion of the car
Friction is ignored, so the force due to friction is assumed negligible
