Ask question

Ask question

All categories

3 years ago

9

Which factors affect the strength of the electric force between two objects

2 answers:

Margaret [11]3 years ago

8 0

<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>

Nikitich [7]3 years ago

5 0

The answer is C I just took the the test

You might be interested in

Why are light-years more convenient than miles, kilometers, or astronomical units (au) for measuring the distances to stars and

Tomtit [17]

Answer:

sEE BELOW

Explanation:

Well.....because the numbers are 'astronomical'....meaning VERY, VERY , VERY LARGE

5 0

1 year ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A race car is accelerating at 5m/s and then speeds up to a final velocity of 12 m/s if the race car driver finished this let of

noname [10]

Answer:

the acceleration of the car is 1.167 m/s²

Explanation:

Given;

initial velocity of the race car, u = 5 m/s

final velocity of the race car, v = 12 m/s

time to finish the race, t = 6 s

The acceleration of the car is calculated as;

a = (v - u) / t

a = (12 - 5) / (6)

a = 1.167 m/s²

Therefore, the acceleration of the car is 1.167 m/s²

3 0

3 years ago

Read 2 more answers

Arlecino [84]

Answer:

A) Diagonals of a parallelogram bisect each other

6 0

3 years ago

Need this done by today! Pls help:(

-Dominant- [34]

Initial Velocity = 0
Final Velocity=3
Duration of time=0.5
Acceleration=6
I hope that helped. Good luck!

5 0

3 years ago