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3 years ago

The mass of the sun is 1.99x10^30 kg . Jupiter is 7.79x10^8 km away from the sun and the mass of 1.90x10^27 kg

Physics

1 answer:

maks197457 [2]3 years ago

6 0

Question is missing:

"What is the gravitational force between the Sun and Jupiter?"

Answer:

$4.16\cdot 10^{23} N$

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the sun

$m_2 = 1.90\cdot 10^{27} kg$ is the mass of Jupiter

$r=7.79\cdot 10^8 km = 7.79\cdot 10^{11} m$ is their separation

Solving the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(1.90\cdot 10^{27})}{(7.79\cdot 10^{11})^2}=4.16\cdot 10^{23} N$

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A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne

SVETLANKA909090 [29]

Given Information:

Resistance of circular loop = R = 0.235 Ω

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:

Magnetic field = B = ?

Answer:

Magnetic field = 0.00145 T

Explanation:

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

8 0

3 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

Gravity at a height h above the surface of the moon will be given as:

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago

4. Premature wrinkling due to overexposure to the Sun, what am I?

Irina-Kira [14]

Premature Aging

Mark as brainlist

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2 years ago

Name the fundamental units involved in pascal

gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

$pressure = \frac{f}{a}$

Now, As per the formula the units are:

kg, m and s^2.

Hope it helps...

5 0

3 years ago