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viktelen [127]
3 years ago
12

A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

r the bullet to hit the target ?
Physics
1 answer:
Alex3 years ago
5 0

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

   Speed = \frac{distance}{time taken}

   Since time is unknown, we make it the subject of the expression;

   time = \frac{distance }{speed} = \frac{400}{600}

   time = 0.67s

Learn more:

Speed brainly.com/question/10048445

#learnwithBrainly

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The heat of fusion for water is the amount of energy needed for water to <span>melt. (c)</span>
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When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di
AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

T = 2\pi \sqrt{\frac{m}{k}}

Where,

m = Mass

k = Spring constant

T = 2\pi \sqrt{\frac{0.35}{28.58}}

T = 0.685s

Therefore the period of the oscillations is 0.685s

4 0
3 years ago
Abdominal breathing is a condition in which only the inferior half of the lungs can be seen expanding and contracting with each
lilavasa [31]

Answer:

The correct answer is - Damage to the nerves that control the diaphragm.

Explanation:

Abdominal breathing is a condition in which inferior half of the lungs can be seen relaxing or contracting and expanding with the breath. This condition occurs due to the various conditions that lead to the respiratory.

It is cause due to the damage to nerves that control the diaphragm. The phrenic nerve is one of the nerve of diaphragm initiates in the neck and passes down.

Thus, the correct answer is - Damage to the nerves that control the diaphragm.

The phrenic nerve is a nerve that originates in the neck (C3–C5) and passes down between the lung and heart to reach the diaphragm.

7 0
3 years ago
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.
andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values ​​where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0
3 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

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2 years ago
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