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3 years ago

8

Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

r =vt/2pi r =act/pi

2 answers:

jek_recluse [69]3 years ago

7 0

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

$\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}$

Learn more about circular motion:

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11Alexandr11 [23.1K]3 years ago

6 0

Answer:

a and d

Explanation:

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When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Oduvanchick [21]

First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
$a = r \times ( \frac{1}{n1} - \frac{1}{n2} )$
We insert the values

$a = 2.18 \times {10}^{ - 18} \times ( \frac{1}{ {1}^{2} } - \frac{1}{ {6}^{2} } )$
$= 2.12 \times {10}^{ - 18}$
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
$wavelength = \frac{h \times c}{a}$
Finally we insert the values
$\frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2.12 \times {10}^{ - 18} } = 9.376 \times {10}^{ - 8}$
Which is the same as 93.8 nm

3 0

3 years ago

In what part of the plant is glucose suger made?

Eduardwww [97]

$\large \mid \underline {\bf {{{\color{navy}{Leaf \: \: \: Chloroplast \: ...}}}}} \mid$

<h2>☛ More Information :</h2>

Green plants manufacture glucose through a process that requires light, known as photosynthesis.

Glucose is stored in the form of starch in plants.

5 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

4 years ago

In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

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4 0

1 year ago

Suzette had prepared the graph below to add to her lab

Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0

3 years ago

Read 2 more answers