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dimulka [17.4K]

3 years ago

8

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

n zero.equal to zero.c. equal to the average of the two distances times the charges.d. equal to the net electric field.

1 answer:

Marizza181 [45]3 years ago

8 0

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

You might be interested in

What's the average speed of an object moving 3790 meters in 249 s.

NISA [10]

Heya!!

For calculate velocity, lets applicate formula

$\boxed{d = v * t}$

<u>Δ Being Δ</u>

d = Distance = 3790 m

t = Time = 249 s

v = Velocity = ?

⇒ Let's replace according the formula and clear "v":

$\boxed{v= 3790\ m / 249\ s}$

⇒ Resolving

$\boxed{ v = 15,22 \ m/s }$

Result:

The velocity is <u>15,22 meters per second.</u>

Good Luck!!

4 0

2 years ago

The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired

My name is Ann [436]

Answer:

Relation between initial speed of bullet and height h is given as

$v = \frac{m + M}{m}\sqrt{2gh}$

Explanation:

As we know that system of block and bullet swings up to height h after collision

So we have

$(m + M)gh = \frac{1}{2}(m + M)v_1^2$

so we have

$v_1 = \sqrt{2gh}$

so speed of the block + bullet just after the impact is given by above equation

Now we also know that there is no force on the system of bullet + block in the direction of motion

So we can use momentum conservation

$mv = (m + M)v_1$

now we have

$v = \frac{m + M}{m}\sqrt{2gh}$

5 0

3 years ago

Which is an example of kinetic energy?

zysi [14]

It’s D because kinetic energy is the energy of motion

8 0

3 years ago

Read 2 more answers

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

<u>Where:</u>

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

7 0

3 years ago

A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th

Gala2k [10]

Answer:

$F_a_v_g=7093333.33N*s$

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

$F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}$

Where:

$m=mass\hspace{3}of\hspace{3}the\hspace{3}object$

$v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval$

$v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.$

$t_2=final\hspace{3}time$

$t_1=initial\hspace{3}time$

Asumming v1=0 and t1=0:

$F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s$

8 0

3 years ago