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DerKrebs [107]
2 years ago
8

What would happen to a 100N gravitational force if both masses were doubled and the radius were

Physics
2 answers:
grin007 [14]2 years ago
7 0

Answer:

They are the only joints that can do 360 degrees and rotate with their own axis. But, because of its free-moving, it is prone to any dislocation compared to other movable joints.

Explanation:

Alik [6]2 years ago
6 0

The force winds up unchanged, at 100 N.

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Diffraction is observable when the opening is ______ than the wavelength.
masya89 [10]
Diffraction is observable when the is smaller than the wavelength 
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3 years ago
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cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your
sp2606 [1]

We have that the time in seconds, minutes, and hours is

t=1.083*10^{-4}s

T_{min}=1.805*10^{-6}min

T_{hours}=3.0083*10^{-8}hours

From the Question we are told that

Velocity v=13.0m/s

Distance d=120 km

Generally the equation for the Time  is mathematically given as

t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s

Therefore

T_{min}=1.083*10^{-4}s/60

T_{min}=1.805*10^{-6}min

And

T_{hours}=T_{min}/60

T_{hours}=3.0083*10^{-8}hours

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4 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
What should the Architect do to ensure Field-Level Security is enforced on a custom Visualforce page using the Standard Lead Con
OLEGan [10]

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The Architect should Use the expression {!$FieldType.lead.accessible}  within the Visualforce page.

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3 years ago
What is a disadvantage of using moving water to produce electricity
klemol [59]
Most of the time the towns get flooded and many people die and lose their homes and their jobs, businesses by the floods and the natural environment is destroyed ,air pollution is produced, the ecosystem could be disrupted and dams are known to fall over and dams are really expensive to build, The dams also cause geological damage, example earthquakes and spills happen.
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