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andre [41]
3 years ago
11

A wave moving along a rope has a

Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

V = wavelength * frequency = 1.5 * 5.5 = 8.25 m/s

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A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C
lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0
3 years ago
Which title best fits Rahma's note
Semenov [28]

Answer:

diffusion is the answer.

4 0
3 years ago
Read 2 more answers
What is the period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds
olga2289 [7]
The correct answer for this question is this one: C) 2.5s. T<span>he period and frequency of a water wave if 4.0 complete waves pass a fixed point in 10 seconds is that 2.5 s
</span>
Here are the following choices:
<span>A) 0.25s 
B) 0.40s 
C) 2.5s 
D) 4.0s</span>
7 0
3 years ago
Kepler found that the orbits of the planets were _____. circular amorphous oval elliptical
vovangra [49]
<span>Kepler found that the orbits of the planets were elliptical. 
His work was so convincing that to this day, they still are. </span>
7 0
3 years ago
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
alexdok [17]

Actual displacement that he required to move

d_1 = 5.4 km towards North

Displacement that he moved due to snow is

d_2 = 8.1 km at 47 degree North of East

now in vector component form we can say

d_1 = 5.4 \hat j

d_2 = 8.1 cos47 \hat i + 8.1 sin47 \hat j

d_2 = 5.52 \hat i + 5.92 \hat j

now the displacement that is more required to reach the destination is given as

d = d_1 - d_2

d = 5.4\hat j - (5.52 \hat i + 5.92\hat j)

d = -5.52 \hat i - 0.52 \hat j

so the magnitude of the displacement is given as

d = \sqrt{5.52^2 + 0.52^2}

d = 5.54 km

its direction is given as

\theta = tan^{-1}\frac{0.52}{5.52} = 5.38 degree

so it is 5.54 km towards 5.38 degree North of West.

4 0
3 years ago
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