In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.
The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.
Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.
We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.
Torque is (force) x (distance from the hinge).
4.9 N-m = (force) x (0.89 m)
Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)
= 5.506 N .
Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock: 
Pebble: 
(b)Calculate the magnitude of the acceleration of each object when released.
Rock: 
Pebble:
Explanation:
The universal law of gravitation is defined as:
(1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
<em>Case for the rock </em>
<em>:</em>
m1 will be equal to the mass of the Earth 
and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth 
.
Newton's second law can be used to know the acceleration.
(2)
<em>Case for the pebble </em>
<em>:</em>
We know that the measure of an incident ray is: α 1 = 40°.
The index of refraction:
- for the air : n 1 = 1.00,
- for the water: n 2 = 1.33
Snell`s Law of Refraction :
n 1 · sin α 1 = n 2 · sin α 2
sin α 2 = n 1 · sin α 1 / n 2 =
= 1.00 · sin 40° / 1.33 = 0.64278 / 1.33 = 0.4833
α 2 = sin ^(-1) 0.4833
α 2 = 28.9 °
Answer: The angle relative to the water`s surface of the rays when beneath the surface is 28.9°.
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Answer:
15.71 m/s
Explanation:
We are given;
Time; t = 0.2 s
Radius; r = 0.5 m
The circumference will give us the distance covered.
Formula for circumference is 2πr
Thus; Distance = 2πr = 2 × π × 0.5 = π
Linear speed = distance/time = π/0.2 = 15.71 m/s
