A - the amount of time needed to travel the distance

The formula for speed is distance/time and the unit is m/s. Therefore you divide the distance travelled by the time it took to travel it.

6 0

3 years ago

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Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i

shepuryov [24]

<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

$m_{1}$ and $m_{2}$ will be now $2m_{1}$ and $2m_{2}$

$r$ will be now $2r$

Let's rewrite the equation (1) with this new values:

$F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}$ (2)

Solving and simplifying:

$F=4G\frac{m_{1}2m_{2}}{4r^2}$

$F=G\frac{m_{1}m_{2}}{r^2}$ (3)

As we can see, equation (3) is the same as equation (1).

So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>

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3 years ago

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How much voltage is required to run 0.42 A of current through a 150

igomit [66]

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

$R=150 \Omega$ is the resistance

Substituting into the equation, we find the voltage needed:

$V=(0.42)(150)=63 V$

Learn more about voltage and current:

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#LearnwithBrainly

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3 years ago

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Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The triangle’s center of mass is midwa

Helga [31]

Answer:Twice of given mass

Explanation:

Given

Two Particles of Equal mass placed at the base of an equilateral Triangle

let mass of two equal masses be m and third mass be m'

Taking one of the masses at origin

Therefore co-ordinates of first mass be (0,0)

Co-ordinates of other equal mass is (a,0)

if a is the length of triangle

co-ordinates of final mass $(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Given its center of mass is at midway between base and third vertex therefore

$x_{cm},y_{cm}=\frac{a}{2},\frac{\sqrt{3}a}{4}$

$y_{cm}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

$\frac{\sqrt{3}a}{4}=\frac{m\cdot 0+m\cdot 0+m'\cdot \frac{\sqrt{3}a}{2}}{m+m+m'}$

$2m+m'=4\times (\frac{m'}{2})$

$2m+m'=2m'$

$m'=2m$

8 0

3 years ago