Ask question

Ask question

All categories

olya-2409 [2.1K]

3 years ago

14

An object with a mass of 70 kilograms is supported at a height 8 meters above the ground. What's the potential energy of the obj

ect with respect to the ground? A. 3,430 J B. 2,058 J C. 5,488 J D. 560 J

1 answer:

alexira [117]3 years ago

5 0

Ep= mgh

70 x 9.8 x 8

Ep= 5,488J

You might be interested in

A stone tied at one end of the string is whirled in vertical circle, where is the tension in the string be maximum and minimum?

Ber [7]

<h3>answer</h3>

The minimum velocity required for a object to rotate in a verticle plane is v

1

=

5

gr.

So, at bottom point, T

max

=mv

1

2

/r+mg=6mg.

At top-most point velocity is v

2

=gr.

So, T

min

=mv

2

2

/r−mg=0

So T

max

−T

min

=6mg=2×g⇒m=1/3 kg

4 0

3 years ago

Two forces of magnitudes 4.0 Newtons and 3.0 Newtons pull on a box. The forces make an angle of 40° with each other. What is the

leonid [27]

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

$F_{1x} = 4.0 N\\F_{1y} = 0$

$F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N$

So the resultant has components

$F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N$

So the magnitude of the resultant is

$F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N$

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.

3 0

3 years ago

How much positive and negative charges is there in a cup of water?

Rina8888 [55]

In a cup of water it is a positive and negative charge of zero

3 0

3 years ago

A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont

PIT_PIT [208]

Answer:

$v' = 2.83 m/s$

Explanation:

Velocity of wave in stretched string is given by the formula

$v = \sqrt{\frac{T}{\mu}}$

here we know that

T = 4 N

also we know that linear mass density is given as

$\mu = 1 kg/m$

so we have

$v = \sqrt{\frac{4}{1}} = 2 m/s$

now the tension in the string is double

so the velocity is given as

$v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s$

$v' = 2.83 m/s$

4 0

3 years ago

Helppppp plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz!

Fynjy0 [20]

C real,inverted and smaller than the object

6 0

3 years ago