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Lorico [155]
3 years ago
14

Which of the following is an open circuit?

Physics
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

can u sent a pic or something

Explanation:

plz needed a clear info

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Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da
FrozenT [24]

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light \lambda =550nm=550\times 10^{-9}m

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So x=3\times 10^{-3}m

We have to find the width of the slit

For the first order wavelength is equal to \lambda =\frac{x}{D}\times a, here a width of slit

So a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm

So width of the slit will be equal to 1.47 mm

5 0
3 years ago
Find the total volume of the material used to making the cylinder​
gladu [14]

Answer:

Hey mate

Explanation:

Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.

pls mark me as brainliest!!

4 0
2 years ago
If the weight of a person is 500 newton what is his mass on the earth ?
miskamm [114]

Answer:

The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

f= Force (of gravity)=500N

g=acceleration of gravity=1.6m/s^2

m=mass=312kg

m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg

his mass is 312kg

8 0
2 years ago
A smith needs to melt 0.0500 kg of gold at 21.0°C. How much heat must be added? (Remember, she has to heat it to the melting poi
baherus [9]

Answer:

9704.6 J

Explanation:

Total Thermal Energy

= Energy required to bring gold to melting point + Energy required to change the state of gold from solid to liquid

= mcT + <em>m</em><em>l</em><em>f</em><em> </em><em> </em>[bolded is energy used to bring gold to melting point, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em><em>i</em><em>s</em><em> </em><em>s</em><em>t</em><em>a</em><em>t</em><em>e</em><em> </em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>g</em><em>o</em><em>l</em><em>d</em><em> </em><em>f</em><em>r</em><em>o</em><em>m</em><em> </em><em>s</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>o</em><em> </em><em>l</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>]</em>

= (0.0500)(126)(1063 - 21) + <em>(</em><em>0</em><em>.</em><em>0</em><em>5</em><em>0</em><em>0</em><em>)</em><em>(</em><em>6</em><em>.</em><em>2</em><em>8</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em>^</em><em>4</em><em>)</em>

= <u>9704.6</u><u> </u><u>J</u>

4 0
3 years ago
A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field  is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

\Delta A = \frac{1}{2} \Delta \theta R^2

The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

=B\frac{\Delta  A}{\Delta  t}

B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

so, the induce EMF will be

E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V

6 0
3 years ago
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