Hi There,
This is False.
Hope this helped!
Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).
It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:
Potential energy shifts:
Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.
This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
The Hubble Space Telescope is a joint ESA/NASA project and was launched in 1990 by the Space Shuttle mission STS-31 into a low-Earth orbit 569 km above the ground. During its lifetime Hubble has become one of the most important science projects ever. Hope this helps! ~ Autumn :)
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
solid state has <u>the </u><u>most</u> intermolecular force of attraction.
