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3 years ago

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Which astronomer proved that the planets of our solar system move in elliptical orbits? A. Kepler B. Copernicus C. Ptolemy D. Ga

lileo

2 answers:

daser333 [38]3 years ago

5 0

The answer is A) kelper

Ierofanga [76]3 years ago

3 0

The answer is Kepler

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How does the law of conservation of energy apply to machines?

trasher [3.6K]

The answer is letter C

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3 years ago

Read 2 more answers

Mice21 [21]

Answer:

A and A

Explanation:

see paper for work! :)

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2 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

2 years ago

an asteroid whose mass is 2.0 x 10^-4 times the mass of earth revolves in a circular orbit around the sun at a distance that is

Xelga [282]

Answer:

0.0002

Explanation:

First you have to solve exponents first, so you solve 10 to the power of negative 4 which is 0.0001 and now you have 2.0x0.0001 and that equals 0.0002

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3 years ago

A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed $v_1=8 m/s$

initial angular velocity $\omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s$

Speed after 4.1 s is $v_2=2.2 m/s$

$\omega _2=\frac{2.2}{0.6}=3.66 rad/s$

using $\omega _2=\omega _1+\alpha t$

where $\alpha$ is angular acceleration

$3.66=13.33+\alpha \cdot 4.1$

$\alpha =-2.37 rad/s^2$ i.e. clockwise

(b)angular displacement

$\theta =\omega _1t+\frac{\alpha t^2}{2}$

$\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}$

$\theta =54.66-19.75$

$\theta =34.91 rad$

3 0

3 years ago