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bagirrra123 [75]
3 years ago
11

A positively charged rod is brought close to one end of an uncharged metal rod but does not actually touch it. What type of char

ge does the end of the metal rod closest to the positively charged rod acquire?
Physics
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

The end of the neutral rod which is the closest part to the charged rod would acquire a negative charge.

Explanation:

One of the rods is positively charged and one of them is neutral.

And the important part is that <u>they do not touch one another</u>, but get close to each other.

In this case, the end of the neutral rod which is the closest part to the charged rod would acquire a negative charge. This is because of the Coulomb's Law. The opposite charges exert an attractive force to each other. The positive charges attract the negative charges on the neutral rod.

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What happens to the volume of a loaf of bread that is squeezed? The mass? The density?
nalin [4]

Explanation:

The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.

I hope this helped!

Thanks!

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3 0
3 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
W=- \frac{1}{2}kx_f^2
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
8 0
3 years ago
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
5 0
3 years ago
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