Answer:
a) U = 735 J
, b) U = 125.7 J
, c) U = 0 J
Explanation:
The gravitational power energy is
U = mg y - mg y₀
The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part
a) Rope is horizontal
The height in this case is the same length of the rope
y = 2.10 m
w = mg = 350 N
U = 350 2.10
U = 735 J
b) when the angle is 34º
y = L - L cos 34
y = L (1- cos34)
y = 2.10 (1- cos 34)
y = 0.359 m
U = 350 0.359
U = 125.7 J
c) in this case this point coincides with the reference system
y = 0
U = 0 J
Answer:
The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.
Explanation:
This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.
Answer:
Friction, normal force, and weight
Explanation:
If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.
Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)
Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.
Answer:
Newton's law of inertia - His first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... This is the first part cited in Newton's first law; "there is no net force on the airplane and it travels at a constant velocity in a straight line."
Newton's law of acceleration - "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."
Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."
Hope this helps! god bless :)
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Answer:
ΔU = 5.21 × 10^(10) J
Explanation:
We are given;
Mass of object; m = 1040 kg
To solve this, we will use the formula for potential energy which is;
U = -GMm/r
But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.
Thus;
ΔU = -GMm((1/r_f) - (1/r_i))
Where;
M is mass of earth = 5.98 × 10^(24) kg
r_f is final radius
r_i is initial radius
G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Since, it's moving to altitude four times the Earth's radius, it means that;
r_i = R_e
r_f = R_e + 4R_e = 5R_e
Where R_e is radius of earth = 6371 × 10³ m
Thus;
ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)
× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))
ΔU = 5.21 × 10^(10) J
