All categories

3 years ago

What is the net force when forces applied are 300 N to the right and 100 N to the right

Physics

1 answer:

labwork [276]3 years ago

8 0

Answer:

400 Newtons to the right.

Explanation:

You have 300 Newtons that are being applied to the right and you also 100 Newtons to the right. When calculating net force with the forces that go the same direction, you add them. 300 plus 100 is 400. Therefore, it is 400 Newtons or N to the right. Hope this helps!

You might be interested in

A toy helicopter takes off and moves 6 m up and then 1 m back down. What

klio [65]

the answer is 5 m up AKA "B"

6 0

4 years ago

9. Which of the following statements is true about scientific theories?

AleksAgata [21]

I think the correct answer from the choices listed above is option D. Scientific theories summarize patterns found in nature. Although, the statement scientific theories are never proven is somewhat true. They are either disproved or they are never disproved. Hope this answers the question.

6 0

3 years ago

Read 2 more answers

The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of in

Marysya12 [62]

We know, F = k * q₁ * q₂ / r²

Substitute the known values,
F = 9 * 10⁹ * 5 * 7 / (1.2)²

F = 315 * 10⁹ / 1.44

F = 218.75 * 10⁹ N

F = 2.1875 * 10¹¹ N [ Final Answer ]

Hope this helps!

5 0

4 years ago

The Bohr model of the hydrogen atom pictures the electron as a tiny particle moving in a circular orbit about a stationary proto

igor_vitrenko [27]

a) The angular velocity of the electron is $4.12\cdot 10^{16} rad/s$

b) The number of revolutions per second is $6.54\cdot 10^{15}$ rev/s

c) The centripetal acceleration of the electron is $8.98\cdot 10^{22} m/s^2$

Explanation:

This is a problem of uniform circular motion: in fact, the electron orbits around the proton in a uniform circular motion.

The angular velocity of an object in uniform circular motion is given by

$\omega = \frac{v}{r}$

where

v is the linear speed

r is the radius of the trajectory

For the electron orbiting around the proton, we have

$v=2.18 \cdot 10^6 m/s$

$r=5.29\cdot 10^{-11} m$

Therefore, the angular velocity is

$\omega=\frac{2.18\cdot 10^6}{5.29\cdot 10^{-11}}=4.12\cdot 10^{16} rad/s$

The period of revolution of the electron is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = 4.12\cdot 10^{16}rad/s$ is the angular velocity

Substituting,

$T=\frac{2\pi}{4.12\cdot 10^{16}}=1.53\cdot 10^{-16}s$

The period is the time the electron takes to make one complete orbit around the proton; therefore, the number of revolutions of the electrons in one second is:

$f=\frac{1}{T}=\frac{1}{1.53\cdot 10^{-16}}=6.54\cdot 10^{15}$ rev/s