Ask question

Ask question

All categories

2 years ago

14

A 2N and an 6N force pull on an object to the right and a 4N force pulls on the object to the left. If the object has a mass of

0.25 kg what is its acceleration?

1 answer:

wolverine [178]2 years ago

3 0

Answer:

$16m/s^{2}$

Explanation:

Please ask if you have more questions!

You might be interested in

A science teacher who was teaching a unit on minerals modeled mineral formation for the class. She heated granular sugar and a v

Olenka [21]

Answer:

Magnetite and quartz

Explanation:

5 0

3 years ago

Read 2 more answers

A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal

sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

6 0

2 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago

A long straight wire has fixed negative charge with a linear charge density of magnitude 4.6 nC/m. The wire is to be enclosed by

Semmy [17]

Answer: -33.3 * 10^9 C/m^2( nC/m^2)

Explanation: In order to solve this problem we have to use the gaussian law, the we have:

Eoutside =0 so teh Q inside==

the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.

Then we simplify and get

σ= -4.6/(2*π*b)= -33.3 nC/m^2

7 0

2 years ago

Find the gravitational force between the

love history [14]

Answer:

The process is given in the pic.

I have taken the average masses so u substitute the values and solve hope it will help :)❤

7 0

3 years ago