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Nitella [24]
3 years ago
7

If the distance between two objects increased, which of the following things could keep the force of gravity between the two obj

ects the same?
a. increasing the mass of one of the objects.
b. decreasing the mass of one of the objects.
c. increasing the velocity of one of the objects.
d. decreasing the velocity of one of the objects.
Physics
1 answer:
AURORKA [14]3 years ago
7 0
The answer A hope this helped

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Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o
VikaD [51]

Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

6 0
3 years ago
What is the force exerted by a solid surface that opposes gravity?
jonny [76]

Electromagnetic force between the molecules! 
6 0
3 years ago
Read 2 more answers
What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having
faltersainse [42]

Answer:

The wavelength of light is 4.53\times10^{-7}\ m

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

d \sin\theta=n\lambda.....(I)

The separation of the slits

d = \dfrac{1}{N}

d=\dfrac{1}{5000}

d=2\times10^{-6}\ m

Now put the value in equation (I)

2\times10^{-6}\sin13.1^{\circ}=\lambda

Here, n = 1

\lambda=4.53\times10^{-7}\ m

Hence, The wavelength of light is 4.53\times10^{-7}\ m

7 0
3 years ago
In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the
MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v

where:

m_1 = 2.5 kg is the mass of the ball

u_1 = 7.5 m/s is the initial velocity of the ball

m_2 = 70 kg is the mass of the man

u_2 = 0 is the initial velocity of the man

v is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s

So, the man and the ball slides on the ice at 0.259 m/s.

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3 0
3 years ago
suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain
IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

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5 0
3 years ago
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