I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy
Answer:
1.59 seconds
12.3 meters
but if you are wise you will read the entire answer.
Explanation:
This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.
Time
<u>Given</u>
a = 0 This is the critical point. There is no horizontal acceleration.
d = 20 m
v = 12.6 m/s
<u>Formula</u>
d = vi * t + 1/2at^2
<u>Solution</u>
Since the acceleration is 0, the formula reduces to
d = vi * t
20 = 12.6 * t
t = 20 / 12.6
t = 1.59 seconds.
It takes 1.59 seconds to hit the ground
Height of the building
<u>Givens</u>
t = 1.59 sec
vi = 0 Another critical point. The beginning speed vertically is 0
a = 9.8 m/s^2 The acceleration is vertical.
<u>Formula</u>
d = vi*t + 1/2 a t^2
<u>Solution</u>
d = 1/2 a*t^2
d = 1/2 * 9.8 * 1.59^2
d = 12.3 meters.
The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)
The other vi is a vertical speed. It is 0.
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
