Ask question

Ask question

All categories

3 years ago

6

Oil ________.

1 answer:

Kazeer [188]3 years ago

5 0

Oil must be mined from underground.

You might be interested in

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

4 years ago

Based on what you know about how light travels, explain why

Mkey [24]

Answer:

you can't see your back because your neck can only move in fixed direction

8 0

3 years ago

Read 2 more answers

A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is oriented

oee [108]

Answer:

Magnetic flux, $\phi=2.22\times 10^{-3}\ Wb$

Explanation:

It is given that,

Magnitude of the magnetic field, B = 0.078 T

Radius of circular loop, r = 0.1 m

The field is oriented at an angle of θ = 25° with respect to the normal to the surface. The magnetic flux through the surface is given by :

$\phi=BA\ cos\theta$

$\phi=0.078\times \pi \times (0.1)^2\ cos(25)$

$\phi=0.00222\ Wb$

or

$\phi=2.22\times 10^{-3}\ Wb$

So, the magnitude of magnetic flux through the surface is $2.22\times 10^{-3}\ Wb$ . Hence, this is the required solution.

4 0

4 years ago

What is the primary function of the small intestines

ikadub [295]

Digestion and Absorbtion

7 0

3 years ago

The U.S. Navy has long proposed the construction of extremely low frequency (ELF waves) communications systems; such waves could

Artyom0805 [142]

Answer:

length of a quarter wavelength antenna for transmitter elf waves are zero

Explanation:

Find the length of a quarter wavelength antenna for transmitter elf waves

Given information

frequency (ELF waves) → 77Hz

Speed of light → 3 $\times$ $10^8$ m/s

$v=\lambda f-------------1$

where

$\lambda\\$ → wavelength of the antenna

f → frequency of the ELF waves

$v$ → speed of light

so rewrite the equation in $\lambda$ from

$\lambda=\dfrac{v}{f}$

Substitute $3\times10^8m/s$ = $v$

Substitute 75Hz = f

$\lambda=\dfrac{3\times10^8m/s}{77Hz}$

$=3.9\times 10^6$ m

since we get this value,but this size of antenna mount on submarine is not possible.

so $\lambda$ is zero

6 0

4 years ago