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gayaneshka [121]
3 years ago
11

A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do

ne to push this particle towards the body, what form of energy will the particle have?
a. mechanical energy
b. electric potential energy
c. heat energy
d. kinetic energy
Physics
1 answer:
ollegr [7]3 years ago
8 0

Answer:

b. electric potential energy.

Explanation:

The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point
Mandarinka [93]

Answer:

F=9.09\times 10^{-9} N

Explanation:

We are given that

Mass of spherical shell,m_1=1900 kg

Mass=m_2=1.80 kg

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

F=G\frac{m_1m_2}{r^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Using the formula

F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}

F=9.09\times 10^{-9} N

Hence,the gravitational force =F=9.09\times 10^{-9} N

6 0
2 years ago
What is the displacement of the runner in 4 laps? *<br> 2 point
Serhud [2]

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

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3 0
2 years ago
A soccer ball is kicked and left
Vedmedyk [2.9K]

Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

v=v_0+at to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

V_y=25sin(45) so

V_y= 18 m/s Now we can use that as the initial upwards velocity in part a:

v=v_0+at and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

V_x=Vcos\theta which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

V_x=25cos45 and

V_x= 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + \frac{1}{2}at^2 where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = 18(1.8)+\frac{1}{2}(-9.8)(1.8)^2 and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

6 0
3 years ago
Dale and Alex push on opposite sides of a couch. Dale pushes to the left with a force of 250 N, while Alex pushes to the right w
bonufazy [111]
F=ma
a=F/m

F=+250N + (-130N)=120N LEFT
('m summing my forces because they are moving about the same axis, and the negative/positive accounts for the arbitrary directions I assign)
m=32kg

F=120N/32kg=3.75m/s^2 to the LEFT

3.75m/s^2 to the LEFT
3 0
3 years ago
Read 2 more answers
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