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Svetach [21]
4 years ago
11

Two mechanical waves can occupy the same space at the same time because waves

Physics
1 answer:
marta [7]4 years ago
6 0
Waves superimpose. You'll get what is called a complex waveform.
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5.4 x 1014Hz
wavelength x frequency = the speed of light
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In what state of matter are particles vibrating and moving at high speeds?
VARVARA [1.3K]

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Gas

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What is my name if it starts with a then m then a then r then y?
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Remember to include your data, equation, and work when solving this problem.
andrezito [222]

Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

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

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

6 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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