The work done by the applied force on the block against the frictional force is 15.75 J.
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Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
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It has to be one continuous column of cloud (air) connected to the ground and in constant rotation.
In soccer, the ball is potential energy. When you kick the ball, it becomes kinetic energy.
Answer:
5 ohms
Explanation:
Given:
EMF of the ideal battery (E) = 60 V
Voltage across the terminals of the battery (V) = 40 V
Current across the terminals (I) = 4 A
Let the internal resistance be 'r'.
Now, we know that, the voltage drop in the battery is given as:
Therefore, the voltage across the terminals of the battery is given as:

Now, rewriting in terms of 'r', we get:

Plug in the given values and solve for 'r'. This gives,

Therefore, the internal resistance of the battery is 5 ohms.
we know the equation for the period of oscillation in SHM is as follows:
T = 2 * pi * sqrt(mass/k)
we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).
since d = v*T, we can say v = d/t = d * f
the final equation, after combining everything, is as follows:
v = d/(2 * pi) * sqrt(k/m)
by plugging everything in
v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))
We find our velocity to be:
v = 6.89 m/s