21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe
1 answer:
The magnitude of the force of friction is 40 N
Explanation:
To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:
- The constant pushing force forward, of magnitude F = 40 N
- The frictional force, acting backward,
Since the two forces are in opposite direction, the equation of motion is
where
m is the mass of the student+scooter
a is the acceleration
However, here the scooter is moving at constant speed: this means that its acceleration is zero, so
a = 0
And therefore,
which means that the magnitude of the force of friction is also equal to 40 N.
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Answer:
Explanation:
Given that,
The position of a particle is given by :
Let us assume we need to find its velocity.
We know that,
So, the velocity of the particle is .
The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.
<h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.
They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.
The given data in the problem is;
q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C
F is the repulsive force = 36 mN =36 ×10⁶ N
d is the distance = 0.70 m
The Coulomb force is found as;
Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.
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The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge C is placed which is the origin.
Let B be the point where the charge C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law, ,
=
k is the Coulomb's law constant.
On substituting the values into the above equation, we get,
Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
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