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jeyben [28]
3 years ago
11

Is the sum of kinetic and gravitational potential energy conserved in a closed system with no additional forces including gravit

y?
Physics
1 answer:
patriot [66]3 years ago
5 0
The acceleration due to gravity changes as we move away from Earth, and the expression for gravitational potential energy must reflect this change. The total energy of a system is the sum of kinetic and gravitational potential energy, and this total energy is conserved in orbital motion.
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Suppose 2.10 C of positive charge is distributed evenly throughout a sphere of 1.30-cm radius. 1) What is the charge per unit vo
Anuta_ua [19.1K]

Answer:

\rho=2.28\times 10^5\ C/m^3

Explanation:

Given that,

Charge, Q = 2.1 C

The radius of sphere, r = 1.3 cm = 0.013 m

We need to find the charge per unit volume for this situation. It can be calculated a follows:

\rho=\dfrac{Q}{\dfrac{4}{3}\pi r^3}\\\\\rho=\dfrac{2.1}{\dfrac{4}{3}\pi \times (0.013)^3}\\\\\rho=2.28\times 10^5\ C/m^3

So, the charge per unit volume is 2.28\times 10^5\ C/m^3.

3 0
3 years ago
So i want to start a dance academy. what do i do
timama [110]

Answer:

get students and a license

5 0
4 years ago
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What type of pipe wrenches are designed for turning and holding where marring is not objectionable?
galina1969 [7]
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5 0
4 years ago
An experiment is performed to test the effect of three different types of water on the growth of a plant. The test is done by us
natima [27]
I think the correct answer would be D. The tap water in the experiment is one the three test conditions of the independent variable, the type of water. The independent variable in a experiment is the one being manipulated or the one being changed. In this case, it is the type of water.
6 0
3 years ago
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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
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