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Greeley [361]
3 years ago
13

Can someone explain to me how to answer this problem in simple ways? (I get really confused when it comes to math) A communicati

ons satellite transmits a radio wave at a frequency of 5 • 109 Hz. What is the signal’s wavelength? Assume the wave travels in a vacuum.
Physics
1 answer:
pishuonlain [190]3 years ago
4 0
When I find a problem like this, I find it helpful to think about what I know and what equation will help me.

The question tells us the frequency of the wave (5 \times 10^{-9}) Hz. We want to work out the wavelength. What equation links these two quantities?

Wave Speed = Frequency x Wavelength
(we know all electromagnetic waves travel at the speed of light in a vacuum)

(3 \times 10^8) = (5 \times 10^{-9}) \times \lambda
and then divide by (5 \times 10^{-9}) to get the wavelength.

Wavelength = 6 \times 10^{16} m<u />
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Priyadarshini means..........​
Blizzard [7]

Answer:

A submission from India says the name Priyadharshini means "Beloved and pleasing to look at" and is of Sanskrit origin. According to a user from India, the name Priyadharshini is of Indian (Sanskrit) origin and means "God gift".

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3 years ago
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When Earth and the Moon are separated by a
Wewaii [24]
     Using the Universal Gratitation Law, we have:

F= \frac{MmG}{d^2}  \\ MmG=2*10^{20}*(3.84*10^8)^2 \\ MmG=29.4912*10^36
 
     Again applying the formula in the new situation, comes:

F= \frac{MmG}{d^2} \\ F= \frac{29.4912*10^36}{(1.92*10^8)^2} \\ \boxed {F=8*10^{20}}

Number 4

If you notice any mistake in my english, please let me know, because i am not native.
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3 years ago
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The amount of gravitational potential energy released as:_________.
nordsb [41]

Answer:

b.it depends on the distance it falls

8 0
3 years ago
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri
fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0
3 years ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
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