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hammer [34]
3 years ago
5

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

s 1.5multiply1011 m. (a) Calculate the magnitude of the gravitational force exerted by the Sun on the Earth. N (b) Calculate the magnitude of the gravitational force exerted by the Earth on the Sun.
Physics
1 answer:
spayn [35]3 years ago
7 0

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = \frac{GMm}{R^2}

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = \frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2} = <em>3.56 x 10^22 N</em>

<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

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Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it  ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to  develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

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3 years ago
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe
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Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀    (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0
3 years ago
x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering
ValentinkaMS [17]

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

2d\sin\theta=n\lambda

For second order, n = 2

\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

4 0
4 years ago
A cubic box has a 0.6 cubic decimetre side. Find the mass of air in the box in gram. (1 Liter of air has a mass of 1.3 g).
morpeh [17]

Answer:

The mass of air in the box is 0.78 grams.

Explanation:

A litre equals a cubic decimetre. By definition of density (\rho), in grams per litre, we calculate the mass of air inside the cube (m), in grams:

m = \rho \cdot V (1)

Where V is the volume occupied by air within the cube, in litres.

If we know that V = 0.6\,L and \rho = 1.3\,\frac{g}{L}, then the mass of air is:

m = \left(1.3\,\frac{g}{L} \right)\cdot (0.6\,L)

m = 0.78\,g

The mass of air in the box is 0.78 grams.

4 0
3 years ago
While rearranging a dorm room, a student does 310 J of work in moving a desk 2.9 m. What was the magnitude of the applied horizo
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Answer:

The horizontal force is 106.89 N.

Explanation:

Given that,

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F=106.89\ N

Hence, The horizontal force is 106.89 N.

3 0
3 years ago
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