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timofeeve [1]
4 years ago
8

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

satellite to the surface of earth if the satellite is orbiting at a height of 9.6 × 106 meters? (hint: time is distance divided by speed.)
Physics
1 answer:
nlexa [21]4 years ago
6 0
<span>3.2x10^-2 seconds (0.032 seconds)
   This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

   9.6 x 10^6 m / 3x10^8 m/s

   First, divide the significands like you would normally.
 9.6 / 3 = 3.2

   And subtract the exponent. So
 6 - 8 = -2

   So the answer is 3.2 x 10^-2
 And since the significand is less than 10 and at least 1, we don't need to normalize it.

   So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>
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A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
Drupady [299]

Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

Current in wire=40 A

Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

Substitute the values in the given formula

B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

B=4.5\times 10^{-5} T

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

7 0
3 years ago
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Aleks [24]

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

3 0
3 years ago
If a bus travel 200 km in 45 minutes calculate the speed in kilometre per minute​
erma4kov [3.2K]

Answer:

multiply that and divided by 45

6 0
3 years ago
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Explain how each of the following changes as you travel across the spectrum: wavelength, frequency, and energy
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As you go from Radio waves across to Gamma eats : Wavelength decreases and frequency and energy increase.
4 0
3 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
4 years ago
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