This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.
Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.
<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
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<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>Given data
The angle of inclination of the plane is theta = 30 degree
The applied force in the inclined plane is F = 94 N
The distance moved in the inclined plane is d = 2.30 m
The coefficient of kinetic friction is u_k = 0.280
The free-body diagram of the above configuration is shown below:
Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.
(a)
The expression for the work done by the pushing force is given as:
Substitute the value in the above equation.
Thus, the work done by the pushing force is 216.2 J.
(b)
The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.
The expression for the normal reaction force is given as:
The expression for the mass of the box is given as:
Substitute the value in the above equation.
Thus, the mass of the box is 12.9 kg.
The time when the two players will collide is 0.96 s.
The equation of motion of the two players is given as;
x1 = 0.1 m + (–3.9 m/s )t
x2 = –6.3 m + (2.8 m/s )t
The time when the two players collide, their displacement is equal or the difference in their position will be zero.
Thus, the time when the two players will collide is 0.96 s.
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