Hi there!
We can use the work-energy theorem to solve.
Recall that:

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

Now, we can define work:

Now, plug in the values:

Solve for theta:

Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is

Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e

- inversely proportional to its cross section area i.e

Therefore

ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:


......(1)
Again for tungsten:

........(2)
Given that
and 
Dividing the equation (1) and (2)

[since
and
]



Therefore the ratio of diameter of the copper to that of the tungsten is

Answer:
(7.8) x (9.8 m/s) = 76.44 m/s
during the time he spent falling.
Since his falling speed was zero when he 'stepped' off of the top,
he hit the ground at 76.44 m/s.
That's about 170 miles per hour.
I'll bet he left one serious crater!
I hope this helps too! :D
Explanation:
Answer:
nitrogen
Explanation:
because I also had this in exam and I was correct
Elements that give up electrons easily are called <u>metals.</u>
hope this helps!