Answer:
(a) To draw water from a well we have to pull at the rope.
(b) A charged body attracts an uncharged body towards it.
(c) To move a loaded trolley we have to pull it.
(d) The north pole of a magnet repels the north pole of another magnet.
Explanation:
Just trust me
Answer: I feel that 3 is the answer
Explanation: Let there be 2 objects, A and B
A is at height of 5m whereas B is at height of 15m
so over here let the gravitational potential energy of A be x
and since B is 3 times higher than A B=3x
Since, earth is considered to be the point where gravitational potenial is 0
So hence forth and object 3 times up will have 3 times the gravitational potential energy of A
Answer:
Explanation:
A. The kinetic energy is the same as the initial potential energy:
PE = mgh = (215 N)(2.0 M) = 430 J
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B. The velocity achieved by falling from a height h is given by ...
v = √(2gh)
v = √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)
v ≈ 6.26 m/s
The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
Answer:
The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.
Explanation:
Given that,
Velocity of ship = 2.00 m/s due south
Velocity of boat = 5.60 m/s due north
Angle = 19.0°
We need to calculate the component
The velocity of the ship in term x and y coordinate


The velocity of the boat in term x and y coordinate
For x component,

Put the value into the formula


For y component,

Put the value into the formula


We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat
For x component,

Put the value into the formula


For y component,

Put the value into the formula


Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.