Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :
We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :
At t = 1 s,
Current,
So, the current at t = 1 s is 3 A.
For lowest current,
Hence, this is the required solution.
Answer:
2271.16N/C upward
Explanation:
The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.
First we determine the downward weight using
Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be
To calculate the electric field,
Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.
Hence the magnitude of the electric force is 2271.16N/C and the direction is upward
(a) 0.165 m/s
The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):
The final total momentum is instead:
where
is the mass of the astronaut
is the velocity of the astronaut
is the mass of the capsule
is the velocity of the capsule
Since the total momentum must be conserved, we have
so
Solving the equation for , we find
(negative direction means opposite to the astronaut)
So, the change in speed of the capsule is 0.165 m/s.
(b) 520.8 N
We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:
The change in momentum of the astronaut is
And the duration of the push is
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.
(c) 25.9 J, 390.6 J
The kinetic energy of an object is given by:
where
m is the mass
v is the speed
For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N