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MatroZZZ [7]
3 years ago
10

What are 3 things that can affect gravity?

Physics
1 answer:
den301095 [7]3 years ago
5 0
The force of gravity the masses exert on each other. If one of the masses is doubled , the force of gravity between the objects is doubled. Increases , the force of gravity decreases.
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The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.
max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg ,  L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3  ⇒  I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3  

α = 1.193 rad / s²

ac = α *2    ⇒  ac = 1.193 rad / s²  * 2

ac = 2.86 m / s²

4 0
3 years ago
A comet is a
Juli2301 [7.4K]
The answer is D. Small object made of ice and dust that orbits the Sun and forms a coma as it approaches the Sun.
7 0
3 years ago
What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.
klio [65]
About 40 000 km/h
Here you go:

8 0
3 years ago
Read 2 more answers
I need the correct answer for this question please​
kirill115 [55]
Try C. a downward gravitational force exerted by Earth
7 0
3 years ago
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
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