Ask question

Ask question

All categories

3 years ago

14

A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw

1 answer:

Alona [7]3 years ago

4 0

Answer:a

Explanation:

You might be interested in

What is a period and how does it work

gtnhenbr [62]

A period is a form of grammer , and it is used to end a sentence

7 0

3 years ago

Read 2 more answers

You and your friends are pushing a car with a total of 1080 N of force. You are able to push the car for a distance of 218 m. Ca

harkovskaia [24]

Answer:

How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.

Explanation:

Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).

8 0

3 years ago

The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

katrin2010 [14]

Answer:

Part a)

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

Time period of Earth about Sun is 1 Year

so it is

$T = 1 year = 3.15 \times 10^7 s$

now we know that angular speed of the Earth about Sun is given as

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

$\omega = \frac{2\pi}{3.15 \times 10^7}$

now speed of center of Earth is given as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

$v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})$

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

now transnational kinetic energy of center of Earth is given as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular speed of Earth about its own axis is given as

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

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now moment of inertia of Earth about its own axis

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2$

$I = 9.83 \times 10^{37} kg m^2$

now rotational energy is given as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

Now total kinetic energy is given as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

6 0

3 years ago

What is the distance between the earth and the sun?

quester [9]

Answer:

an average of 92,955,807 miles

Explanation:

7 0

4 years ago

Read 2 more answers

What one unit of cane sugar is made of

Molodets [167]

12 carbon, 22 hydrogen, 11 oxygen

4 0

3 years ago