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frutty [35]
3 years ago
9

Laboratory experiments, observational field studies, and model-building are all examples of different forms of scientific invest

igations. In what way do laboratory experiments primarily differ from other forms of scientific investigations?
A.
A laboratory experiment is the only accepted form of investigation within the scientific community.
B.
Studies about how things behave in nature or studies involving very large objects are best answered through laboratory experiments.
C.
Data can only be generated through laboratory experiments, not other forms of investigation.
D.
Laboratory experiments involve the identification and control of variables.
Physics
1 answer:
Ne4ueva [31]3 years ago
8 0

Answer:

C

Explanation:

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pychu [463]
True, recycling allows products to be reused which saves alot the natural resources that are used.
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4 years ago
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A trombone plays a C3 note. If the speed of sound in air is 343 m/s and the wavelength of this note is
Umnica [9.8K]

The frequency of note C3 is 131 s^{-1}.

<u>Explanation:</u>

Frequency is the measure of repetition of same thing a certain number of times. So frequency is inversely proportional to the wavelength. As wavelength is distance between two successive crests or troughs in a sound wave.

And frequency is the completion of number of cycles in a given time in sound waves. The frequency and wavelength are inversely proportional to each other with velocity of sound being the proportionality constant.

Thus, here the speed of sound is given as 343 m/s, the wavelength of the note is also given as 2.62 m, then frequency will be as follows:

Frequency=\frac{speed of sound}{Wavelength of note}

Thus,

Frequency = \frac{343 m/s}{2.62 m} = 131 s^{-1}

So the frequency of note C3 is 131 s^{-1}.

3 0
3 years ago
How small are the wavelengths of gamma ray radiation?
WARRIOR [948]

Gamma rays are the highest energy EM radiation and typically have energies greater than 100 keV, frequencies greater than 1019 Hz, and wavelengths less than 10 picometers.

3 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.
Dmitry_Shevchenko [17]

Answer:

false statement : b )  For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = -  Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second  option,  the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd  option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

4 0
3 years ago
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