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Drupady [299]
3 years ago
9

Which best describes the purpose of a controlled experiment?

Physics
1 answer:
navik [9.2K]3 years ago
3 0
A controlled experiment is best described as a safe, in depth, and insightful display that helps you understand the purpose of the experiment better
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What is the frequency of a wave if the speed is 24 m/s and the wave is 2 meters
vekshin1
Velocity=frequency(wavelength)
24m/s=f(2m)
24/2=f(2)/2
12Hz=f
4 0
3 years ago
10 basic rules of badminton?​
saw5 [17]

Answer:

The 10 rules of badminton are as follows:

1. A game starts with a coin toss. Whoever wins the toss gets to decide whether they would serve or receive first OR what side of the court they want to be on. The side losing the toss shall then exercise the remaining choice.

2. At no time during the game should the player touch the net, with his racquet or his body.

3. The shuttlecock should not be carried on or come to rest on the racquet.

4. A player should not reach over the net to hit the shuttlecock.

5. A serve must carry cross court (diagonally) to be valid.

6. During the serve, a player should not touch any of the lines of the court, until the server strikes the shuttlecock. During the serve the shuttlecock should always be hit from below the waist.

7. A point is added to a player's score as and when he wins a rally.

8. A player wins a rally when he strikes the shuttlecock and it touches the floor of the opponent's side of the court or when the opponent commits a fault. The most common type of fault is when a player fails to hit the shuttlecock over the net or it lands outside the boundary of the court.

9. Each side can strike the shuttlecock only once before it passes over the net. Once hit, a player can't strike the shuttlecock in a new movement or shot.

10. The shuttlecock hitting the ceiling, is counted as a fault.

Explanation:

8 0
3 years ago
Read 2 more answers
A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0
3 years ago
How do solar panels work with conduction, convection and radiation?
Lady_Fox [76]

Answer:

<em>In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.</em>

Explanation:

3 0
3 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
3 years ago
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