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oksano4ka [1.4K]
3 years ago
10

Why do you think solar panels would work well in space

Physics
2 answers:
julsineya [31]3 years ago
5 0

Answer:

Solar panel equipped, energy transmitting satellites collect high intensity, uninterrupted solar radiation by using giant mirrors to reflect huge amounts of solar rays onto smaller solar collectors.

Explanation:

algol [13]3 years ago
3 0

They work well in space because they receive the full amount of sunlight without any atmosphere or clouds getting in the way. This assumes that objects such as other planets don't cover up the sun. Also, the spacecraft can't be too far away from the sun. If the spacecraft was at Pluto for instance, then the sun is not as effective as compared to at Mars.

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If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
3 years ago
Reaction _______ refers to the tendency of a reaction to occur without the application of outside force or energy to sustain the
babymother [125]
<span>spontaneity is the answer, hope this helps!!</span>
6 0
3 years ago
The blades of a fan running at low speed turn at 26.2 rad/s. When the fan is switched to high speed, the rotation rate increases
Lady bird [3.3K]

Answer: 1.79\ rad/s^2

Explanation:

Given

Initial angular speed is \omega_1=26.2\ rad/s

Final angular speed is \omega_2=36.5\ rad/s

Time period t=5.75\ s

Magnitude of the fan's acceleration is given by

\Rightarrow \alpha=\dfrac{\omega_2-\omega_1}{t}

Insert the values

\Rightarrow \alpha=\dfrac{36.5-26.2}{5.75}\\\\\Rightarrow \alpha=\dfrac{10.3}{5.75}\\\\\Rightarrow \alpha=1.79\ rad/s^2

Thus, fan angular acceleration is 1.79\ rad/s^2

8 0
3 years ago
Read 2 more answers
How do modern scientist describe the make up of matter
lbvjy [14]

The matter is a collection or composition of extremely small particles called "Atoms".

<u>Explanation</u>:

  • The matter is anything that has mass and which utilizes space by having volume.
  • Modern scientists describe the makeup of matter as a collection of tiny substances called atoms.
  • Matter can either be in solid, liquid, or gaseous state and even appear in plasma form.
  • Matter can neither be created nor be destroyed but only transferred from one form to the other.
  • Atoms are composed protons, neutrons and electrons. The protons and neutrons are contained in the nucleus, while the electrons orbit around the nucleus of the atom.
  • Protons are positively charged, neutrons are neutral and electrons are negative.
4 0
3 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
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