Answer:
A force
Explanation:
A push or a pull is an example of a force and can cause an object to speed up, slow down, etc.. Newton's laws tell us that 1- an object will not change its motion unless a force acts on it 2- the force on an object is equal to its mass times its acceleration. 3- The third law states that for every action (force) in nature there is an equal and opposite reaction.. However, forces like gravity and friction can resist movement.
On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s
Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.
Maximum velocity of a road with friction is given by the formula,
v = μRg
where, v is the maximum velocity
μ is the coefficient of static friction
R is the radius of the circle road
g is the acceleration due to gravity
Given,
μ = 0.20
R = 75m
g = 9.8m/s²
On substituting the given values in the above formula,
v = 0.20× 75 ×9.8
v = 147m/s
So, the Maximum velocity of the wet road is 147m/s.
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The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
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Answer:
Option C. 5,000 kg m/s
Explanation:
<u>Linear Momentum on a System of Particles
</u>
Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed
P=mv
The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object


The sum of the momenta of both objects prior to the collision is

