The velocity of the ball and the man is 0.259 m/s
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:
where:
is the mass of the ball
is the initial velocity of the ball
is the mass of the man
is the initial velocity of the man
is the final velocity of the man and the ball after the collision
Re-arranging the equation and substituting the values, we find the final velocity:
So, the man and the ball slides on the ice at 0.259 m/s.
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Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
it cannot be reuse and replaced for long period of time.hope it helps u
The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.
<h3>What is normal force?</h3>
The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.
This force is applied by the solid bodies on each other in order to prevent the passing through each other.
A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.
- The body is gaining the speed, which means there is a change in kinetic energy.
- The change in kinetic energy is equal to the work done.
- The friction force is the product of coefficient of the friction and normal force.
- The friction force for the given case is zero. Thus, the normal force must be equal to the zero.
Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.
Learn more about the normal force here;
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Answer:
0.0002 C.
Explanation:
Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)
Mathematically, charge can be expressed as
Q = CV ................................. Equation 1
Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.
Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.
Substitute into equation 1
Q = 2×10⁻⁶× 100
Q = 2×10⁻⁴ C
Q = 0.0002 C.
The amount of charge accumulated = 0.0002 C
