You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forc
2 answers:
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Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
The electric field strength inside the capacitor is 49880.77 N/C.
Explanation:
Given:
Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m
Separation between the plates (d) = 0.407 mm =
Energy stored in the capacitor (U) =
Assuming the medium to be air.
So, permittivity of space (ε) =
Area of the square plates is given as:
Capacitance of the capacitor is given as:
Now, we know that, the energy stored in a parallel plate capacitor is given as:
Rewriting in terms of 'E', we get:
Now, plug in the given values and solve for 'E'. This gives,
Therefore, the electric field strength inside the capacitor is 49880.77 N/C
The kinetic energy of 2.5 kg ball after collision is 27.09 J.
Answer:
Explanation:
In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.
We know that momentum is the product of mass and velocity acting on any object.
So, the conservation of energy in elastic collision leads to following equation:
Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So
Since initial velocity for M1 ball is zero, then
and
So, on solving all the above equation, we get an equation for velocity and that is
=final velocity of ball with mass 2.5 kg
So kinetic energy will be 1/2 mv2
Kinetic energy of 2.5 kg ball is
So the kinetic energy of 2.5 kg ball after collision is 27.09 J.