All categories

3 years ago

In which instrument of the following the water works as a bulb

Physics

1 answer:

Montano1993 [528]3 years ago

3 0

Answer:

hydrometer

Explanation:

A hydrometer is an instrument used to determine specific gravity. It operates based on the Archimedes principle that a solid body displaces its own weight within a liquid in which it floats. Hydrometers can be divided into two general classes: liquids heavier than water and liquids lighter than water.

You might be interested in

A spectroscope creates a spectrum, or array of colors, based on the light emitted by a star. How are

Ivanshal [37]

In a spectrograph, black lines can be seen going through the array of colors. The pattern of these lines indicate the composition of the star.

Different elements block different parts of the spectrum, resulting in black lines.

6 0

3 years ago

A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long

Airida [17]

Answer:

$0.00027646\ T$

$2.33\times 10^{-5}\ H$

-0.04194 V

Explanation:

$N_2$ = Number of turns in outer solenoid = 330

$N_1$ = Number of turns in inner solenoid = 22

$I_1$ = Current in inner solenoid = 0.14 A

$\dfrac{dI_2}{dt}$ = Rate of change of current = 1800 A/s

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius = 0.0115 m

Magnetic field is given by

$B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T$

The average magnetic flux through each turn of the inner solenoid is $0.00027646\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb$

Mutual inductance is given by

$M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $2.33\times 10^{-5}\ H$

Induced emf is given by

$\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V$

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

3 0

3 years ago

You observe two stars over the course of a year (or more) and find that both stars have measurable parallax. (1 arc second is 1/

Ierofanga [76]

Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

The distance between the Earth and the Sun is 150000000 Km. That distance is also known as an astronomical unit (1AU).

The parallax angle can be defined in the following way:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

$d(pc) = \frac{1}{1}$

$d(pc) = 1 pc$

So, star X is at 1 parsec from Earth.

Case of Star Y ( $p('') = \frac{1}{2}$ ):

$d(pc) = \frac{1}{(\frac{1}{2})}$

$d(pc) = 2 pc$

So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

5 0

3 years ago

Which best describes the way a sound wave is sent through the radio?

Anna35 [415]

Answer:

sound wave-electrical wave-radio wave

4 0

3 years ago

Read 2 more answers

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago