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3 years ago

A milliammeter has an internal resistance of 5ohms and

Physics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

Rs = 0.02008 Ω = 20.08 mΩ

Explanation:

The range of an ammeter can be increased by connecting a small shunt resistance to it in a series combination. This shunt resistance can be calculated by the following formula:

$Rs = \frac{I_gR_g}{I - I_g}$

where,

$R_s$ = value of shunt resistance = ?

$I_g$ = current range of ammeter = 20 mA = 0.02 A

I = Required range of ammeter = 5 A

$R_g$ = Resistance of ammeter = 5 ohms

Therefore,

$R_s = \frac{(0.02\ A)(5\ ohms)}{5\ A-0.02\ A}$

You might be interested in

What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?

Katena32 [7]

Answer:

It would not be possible the cohesion among water molecules by the polar covalent bonding.

Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.

In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule ( $H_{2}O$ ) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.

How is this possible?

By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.

Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.

5 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago

You should be particularly careful a restraining a struggling rabbit because it can

Shtirlitz [24]

You have no options here so I'll just answer. It can cause a rise in heart rate and greatly increases the risk of overheating and even death. If you grab the rabbit too hard, you risk breaking/fracturing a bone or causing other kinds of damage, whether externally or internally, to the rabbit.

6 0

4 years ago

Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the

qwelly [4]

Answer:5.35nm

Explanation:

Consider that 1 inch is = 0.0254m

we have,

1m= 1x10^9 nm

While:

0.0254m = 2.54x10^7nm

1/55 (2.54x10^7) = 4.6181 x 10^5nm

1 day= 24 hrs

= (24x60) when calculating in min

= (24x60x60) calculating in seconds we have:

= 8.64x10⁴sec

In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm

Therefore, the amount by which the hair grows in 1 second will be;

= (4.6181 x 10^5)/(8.64x10^4)

= 5.35nm

The rate of growth will be 5.35nm

7 0

3 years ago