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3 years ago

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Find the change in the force of gravity between two planets when the distance between them is reduced to one-tenth of the origin

al distance.

1 answer:

Fofino [41]3 years ago

7 0

The equation for gravitational force is provided. If you were to put 1/10 times d (distance) than 1/10 would be squared = 1/100. So the gravitational force would be 100 times more.

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ASAP When salt is dissolved in water, the result is a A. gas B. compound C. homogeneous mixture D. heterogeneous mixture

alisha [4.7K]

It’s actually C. homogeneous compound because when they combine you can’t separate them

4 0

3 years ago

Read 2 more answers

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t

Lisa [10]

Explanation:

Let us assume that piece 1 is $m_{1}$ is facing west side. And, piece 2 is $m_{2}$ facing south side.

Let $m_{1} \text{and} m_{2}$ = m and $m_{3}$ = 2m. Hence,

$2mv_{3y} = m(21)$

$v_{3y} = \frac{21}{2}$

= 10.5 m/s

$2mv_{3x} = m(21)$

$v_{3x} = \frac{21}{2}$

= 10.5 m/s

$v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}$

= $\sqrt{(10.5)^{2} + (10.5)^{2}}$

= 14.84 m/s

or, = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

$\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})$

= $tan^{-1}(\frac{10.5}{10.5})$

= $tan^{-1} (45^{o})$

= $45^{o}$

Therefore, the direction of the third piece (degrees north of east) is north of east.

5 0

3 years ago

An uncharged, nonconducting, hollow sphere of radius 10.0cm surrounds a 10.0-μC charge located at the origin of a Cartesian coor

MrMuchimi

The electric flux through the hole is $56.45\ webber$ .

Electric flux is the number of electric field lines cutting through the surface and is measured as surface intregal of electric field over that surface
Mathematically it is given by $\phi_E=E.A \ Nm^2/C$ where E is the electric field and A is the area.
Gauss's law states that electric flux through closed surface is equal to the 1 / ε₀ times the charge enclosed by that surface which is given by Ф = q / ε₀ where q is the central charge and ε₀ is the permittivity of the medium.

It is given , hollow sphere of radius 10.0cm surrounds a 10.0-μC charge.

The whole surface of hollow sphere $= 4\pi r^2$

$= 4\times 3.14\times (10 \times 10^{-2})^2 \\\\= 12.56\times 10^{-2} m^2$

Area of the hole ( both side ) $= 2\times \pi r^2$

$= 2\times 3.14 \times (10^-^3)^2\\= 6.28 \times 10^-^6 m^2$

According to Gauss's theorem, the flow from a particular charge in the center is given by

$\phi= \frac{10\times10^-^6}{8.85\times 10^-^1^2}\\\\\phi=1.13\times10^6$

This flux flows through the surface of the sphere, so the flux per unit area which is given by

$= \frac{ 1.13\times 10^6 }{ 12.56\times 10^-^2} \\\\= 8.99 \times 10^6 \ weber / m^2$

Flux through area of hole is given by :

$= 8.99\times10^6 \times6.28 \times 10^-^6\\ = 56.45 \ weber$

Learn about more electric flux here :

brainly.com/question/26289097

#SPJ4

8 0

1 year ago

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Aneli [31]

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

6 0

3 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

3 years ago