A crane dose 62500 joules of work to lift the boulder distance of 25 meters how much does the boulder way

Physics

2 answers:

Sunny_sXe [5.5K]4 years ago

5 0

Work = force x distance
62,500 joules = boulder's weight x 25 meters
Divide each side by 25 meters:
Boulder's weight = 62,500 joules / 25 meters = 2,500 newtons
That's about 562 pounds

vladimir2022 [97]4 years ago

4 0

Work is equal to force times distance.
-W=F*D or W=FD
Where force is measured in Newtons(N) and distance in meters(m). The unit for work is a newton-meter (aka joule(j))
All you have to do to find how much it weighs is do the opposite. So you would do 62500j/25m. That gives you 2500N, to check it you just multiply 2500N and 25m. Which gives you 62500j.
Hope this helped!

You might be interested in

A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou

lakkis [162]

Answer:

C. while the magnet is moving

Explanation:

Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

8 0

3 years ago

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$