The work to stretch a spring from its rest position is
(1/2) (spring constant) (distance of the stretch)²
E = 1/2 k x² .
You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write
1700 joules = 1/2 k (3m)²
1 joule = 1 newton-meter
1700 N-m = 1/2 k (3m)²
Multiply each side by 2: 3400 N-m = k · 9m²
Divide each side by 9m² k = 3400 N-m / 9m²
= (377 and 7/9) newton per meter
Answer:
Explanation:
Current, I = 6 A
diameter of wire, d = 2.05 mm
number of electrons per unit volume, n = 8.5 x 10^28
If the diameter is doubled,
The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.
According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
<h2>
(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>
(2)
</h2>
Then:
<h2>
(3)
</h2>
Which is the same as:
<h2>
</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
Imagine a car, and imagine you see little arrows below and beneath it, just around it. <span>The path of a particle that is flowing steadily and without turbulence in a fluid past an object.</span>
