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artcher [175]
2 years ago
8

According to Newton's Law of Universal Gravitation, which of the following would cause the attractive force between a planet and

the Sun to be doubled?
A. Doubling the mass of the planet
B. Quadrupling the distance from the sun
C. Doubling the distance from the sun
D. Squaring the mass of the planet oxford
Physics
1 answer:
vladimir1956 [14]2 years ago
4 0

Answer:

it is either a or c

Explanation:

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A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.
Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0
1 year ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
Marysya12 [62]

When air is blown into the open pipe,

L = \frac{nλ}{2}

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ=\frac{2L} {n}

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ=\frac{(2)*(6)}{4} =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= \frac{c}{λ}

⇒f=\frac{344}{3}

≈115 Hz

8 0
2 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
Calculate the weight of a 4.5 kg rabbit.
solniwko [45]
The correct answer is: 13900589.
3 0
2 years ago
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
1 year ago
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