If the distance between two charged particles is doubled, the force between them changes by a factor of

Why is traveling between the stars (by creatures like us) difficult?

mariarad [96]

Answer:

Stars are at large distances that even light takes thousands of years to reach us from there.

for example, alpha centauri is 4300 light years away from earth and it is considered the nearest star to us, this means that light from there takes 4300 years to reach us and with a spaceships that can move with the speed of light is would take us 4300 years to get there which is imposible to live for that long.

hence, it is difficult to move between the stars.

4 0

3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago

If a golf ball and a ping-pong ball both move with the same kinetic energy, which has the greater speed? 1. the golf ball 2. can

schepotkina [342]

Think about it like this, the more mass there is, the faster its going to go. If you took a golf ball and a ping pong ball and you held them each separately, you would notice that the golf ball is heavier. If they move with the same kinetic energy, but the golf ball WEIGHS more, then the golf ball will have the greater speed. If you think about it, the ping pong ball may be taking its time to get to wherever its going.

4 0

4 years ago

The answer and how to do it?? Thanks

denis-greek [22]

Answer:

14 m/s²

Explanation:

Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.

(4N + 10 x 1kg)=(1kg)a

14/1=14, so the acceleration is 14 m/s²

4 0

3 years ago

How strong is naruto i need a in depth answer going off his AP and DC i prefer powerscalers since they do this but anyone can an

marin [14]

Answer:

Naruto is kinda strong

Explanation:

If were just talking about just him I think that he is strong for letting go of his past and moving along with his life. Bu t he also isn't strong if you think about it if Naruto didn't have the nine tails he wouldn't be special so he also is not strong there for his only power really comes from the nine tails.

6 0

3 years ago