Answer:
Mass doesn't change.
Weight is measured based on gravitational pull.
Explanation:
The direction of the centripetal acceleration is towards Saturn
Explanation:
When an object moves in a circular motion, there must be a force that "pulls" the object towards the centre of the circle, keeping it in a circular motion. This force is called centripetal force.
As a consequence, due to the relationship between force and acceleration (Newton's second law), there is also an acceleration that points towards the centre: this acceleration is called centripetal acceleration.
The magnitude of the centripetal acceleration is given by:
where
m is the mass of the object
v is its speed
r is the radius of the circle
Therefore in this situation, the centripetal acceleration points towards the centre of the circle: therefore, towards Saturn, which occupies the centre of the circular trajectory.
Learn more about centripetal acceleration:
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Answer:
Energy = .13 W / m^2 energy of incident energy
N = 3500 Watts / day power needed
N = 3500 Watts (3600 * 24 sec) = .0405 Watts/sec
The problem must mean that one needs 3.5 Kw-days
3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules
150 J/sec-m^2 * .13 = 19.5 J / sec-m^2 usable energy
In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2 usable energy received
Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2
One would need 180 m^2 of solar panels
That's quite a lot of energy
A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.
Answer:
64 J
Explanation:
The potential energy change of the spring ∆U = -W where W = work done by force, F.
Now W = ∫F.dx
So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)
∆U = - ∫-Fdx
= ∫F.dx
Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx
= ∫₀²(40x - 6x²).dx
= ∫₀²(40xdx - 6x²dx)
= ∫₀²(40x²/2 - 6x³/3)
= ∫₀²(20x² - 2x³)
= [20x² - 2x³]₀²
= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)
= [(20(4) - 2(8)) - (0 - 0))
= [80 - 16 - 0]
= 64 J
Complete Question
A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.40 m. The vertical distance from the top of the incline to the bottom is 1.09 m. If , what is the acceleration of the block as it slides down the incline plane
Answer:
The acceleration is
Explanation:
From the question we are told that
The distance from top to bottom of the inclined plane measured along the incline is
The distance from top to bottom of the inclined plane measured along the vertical axis is
According to the SOHCAHTOA rule
=>
substituting values
=>
T
The acceleration of a block on a frictionless inclined plane is mathematically represented as
substituting values