Ohm's Law tells the relationship between voltage, current, and resistance.
It can be written in three different ways, depending on which ones you know,
and which one you want to find.
Here's the one we need:
Resistance = (voltage) divided by (current)
= (120 V) / (0.5 Amp)
= 240 ohms .
Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
Answer:
The correct answer is C. 45.5 lbs.
Explanation:
In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.
The formula for any problem involving a lever is:

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.
The parameter of the formula that you need is F_l:

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=45.5 lbs
Speed (ex: meters/second, miles/hour)
In 0.25h it will move in 22.5 kilometers.