Ask question

Ask question

All categories

3 years ago

10

a horizontal force of 50N acts on a body of mass 20KG initially at rest and displaces it by striking it along the entire 20m pat

h. What has been the work done by said force?

1 answer:

Aneli [31]3 years ago

3 0

Answer:

1000 J

Explanation:

Work = force × distance

W = (50 N) (20 m)

W = 1000 J

You might be interested in

A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?

il63 [147K]

Answer:

$\huge\boxed{\sf P.E. = 39.2\ Joules}$

Explanation:

<u>Given Data:</u>

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

<u>Required:</u>

Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = mgh

<u>Solution:</u>

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

5 0

3 years ago

What is coulomb law

MrRa [10]

Answer:Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Explanation:Coulomb's law, or Coulomb's inverse-square law, is an experimental law of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force

6 0

2 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago

What is the purpose of the wire coil in an electromagnet?

ziro4ka [17]

It is to conduct electricity in the magnet so it has an electric field.

Please BRAINLIEST!

3 0

3 years ago

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 ev, 27 ev, and 48 ev

erastovalidia [21]

Answer:

l=3.5 x 10^-10m

Explanation:

just toook the test got it right good luck!

3 0

2 years ago