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2 years ago

When discussing how charges interact, we can say that

Physics

2 answers:

miss Akunina [59]2 years ago

5 0

Unlike charges are attracted.

Travka [436]2 years ago

3 0

Answer:

Explanation:

According to the properties of charges,

like charges repel each other and the unlike charges attract each other.

If there are two positive or two negative charges, they always repel each other.

If there is one positive or one negative charge, they always attract each other.

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Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th

irinina [24]

I'm not quite sure what happens to Fay so I didn't finish but hope it helps

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2 years ago

Read 2 more answers

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

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2 years ago

A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe

aniked [119]

We are given information:
$m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N$

If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2

Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s

Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s

Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s

7 0

2 years ago

A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1

kati45 [8]

<span>The formulas are, v1d1Â˛ = v2d2Â˛ ........ (1) h = (v2Â˛-v1Â˛)/2g ...... (2) Given that, v1 = 1.71 m/s we assume that the stream has decreased by a factor d2 =0.805d1 then, v1d1Â˛ = v2 (0.805d1)Â˛ cancelled both side d1Â˛ then we get, v1 = v2 (0.805)Â˛ v1 = v2 (0.648025) Sub v1 = 1.71, 1.71 = v2 (0.648025) v2 = 1.71/0.648025 v2 = 2.638787083831642 v2 = 2.64 m/s The vertical distance formula, h = (v2Â˛-v1Â˛)/2g We know that value of gravity constant is 9.8 m/sÂ˛ h = {(2.64)Â˛ - (1.71)Â˛)/2(9.8) h = {(6.9696) - (2.9241)}/19.6 h = (4.0455)/19.6 h = 0.2064030612244898 h = 0.21 cm Therefore, the vertical distance h = 0.21 cm.</span>

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2 years ago

Do as physics instructor fred cauthen does and place a tennis ball close to and above the top of a basketball. drop the balls to

Ksivusya [100]

Answer:

after shock

creating a system for the conservation of the energy of the basketball ball and creating a system for the tennis ball only, the conservation of energy should be applied to each system independently

Explanation:

When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.

Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock

In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.

Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

Em₀ =K = 1/2 m v²

$Em_{f}$ = U = m g h

As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height

m g h = 1/2 m (2v₀)²

h = 2 v₀²/g

To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.

6 0

2 years ago