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damaskus [11]
3 years ago
10

A neutral metal ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to

be attracted to the rod. This is because:____________.
a. the ball becomes negatively charged by induction
b. the ball becomes positively charged by induction
c. the string is not a perfect insulator
d. there is a rearrangement of the electrons in the ball
e. the number of electrons in the ball is more than the number in the rod
Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

d. there is a rearrangement of the electrons in the ball

Explanation:

Inside the neutral metal ball, there are equal no. of positive charges (protons) and negative charges (electrons). Normally, the charges are distributed evenly throughout the ball.

However, when the positively charged insulating rod is brought near, since positive charges and negative charges attract each other, the electrons (-ve charges) in the metal ball moves towards the side nearest to the rod. The metal ball gets attracted to the rod.

a and b are not correct because the rod is insulating, so electrons cannot be transferred between them to induce a net charge in the metal ball. the no. of electrons is unrelated to the attraction between opposite charges , so e is incorrect as well.

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2 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

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As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
The distance from earth to teh sun usb approxximately 93 million miles. a scientist would write that number as
erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as  1.49 x 10⁸ kilometres .)

6 0
3 years ago
plz i need help fast................a motion along a straight line with a constant increase in velocity is ?​
Monica [59]

Answer:

uniform acceleration

Explanation:

The definition for uniform acceleration is:

if an object travels in  a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.

Hope this helps.

Good Luck

8 0
3 years ago
Read 2 more answers
A slice of cheese has a mass of 40 g and a volume of 23 cm^3. What is the density of the cheese in units of g/cm^3 and g/mL?
Mila [183]

The mathematical and proportional relationship between mL and cm ^ 3 said us that 1cm ^ 3 is equivalent to 1mL.

If the density is considered as the amount of mass per unit volume we will have to

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here,

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\rho = \frac{40g}{23cm^3}

\rho = 1.739g/cm^3

As 1mL = 1cm^3 we have that the density in g/mL is,

\rho = 1.739g/mL

6 0
3 years ago
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