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3 years ago

A man weighing 490N on earth weighs 81.7N on the moon.His mass on the moon is kg

2 answers:

6 0

His mass is 50 kg . . . on the Moon, on the Earth, in the capsule rocketing between them, and on Halley's comet if he ever goes there.

Veseljchak [2.6K]3 years ago

6 0

Answer:

50kg

Explanation:

The mass of a man on the Earth and the moon is the same, so we only need to use the 490N weight on Earth.

The formula for mass is m=w/g. Plugging in the values, we can determine that m=490N/9.8m/s^2. (g is gravitational constant, which is approximately 9.8m/s^2 on Earth)

The reason that the unit in the answer is in kg is because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the m/s^2 in the gravity constant, so it leaves kg.

m=50kg

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Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the

Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

$\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}$

The force can be calculated by the formula

$F=k\frac{q1q2}{(r^{\prime})^2}$

Here, k is the constant whose value is

$k=9\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the force will be

$\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}$

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1 year ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

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3 years ago

Clara rushes 30m to a truck then turns and walks back. Total travel is 120s what is her average velocity?

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Taking into account the definition of velocity, Clara's average velocity is 0.5 m/s.

<h3>Definition of velocity</h3>

Velocity is a physical magnitude that relates the displacement of an object, the time it takes to make this change in position and direction. So it is considered a vector magnitude.

In other words, the velocity can be defined as the amount of space traveled per unit of time with which a body moves, considering the direction, and can be calculated using the expression:

velocity= distance traveled÷ time

<h3>Average velocity of Clara</h3>

Clara rushes 30 m to a truck then turns and walks back. Total travel is 120s. Then, you know:

distance traveled= 30m rushing + 30m walking back= 60 m
time= 120 s

Replacing in the definition of velocity:

velocity= 60 m÷ 120 s

Solving:

<u><em>velocity= 0.5 m/s</em></u>

Finally, Clara's average velocity is 0.5 m/s.

Learn more about velocity:

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2 years ago

What is the difference between each distance traveled and displacement travled

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Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

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3 years ago

For Valentine’s Day, Sally received a helium-filled balloon at a party. On returning home she accidentally left the balloon in t

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Answer:

If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.

As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.

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3 years ago