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3 years ago

Name the device of measurement and write its used or its function?

Physics

1 answer:

labwork [276]3 years ago

4 0

Answer:

There is a lot of instruments used for measurement, may I ask which one are you referring to?

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

What happens when a negatively charged object A is brought near a neutral object B?

castortr0y [4]

Answer: Option B

Explanation : When a negatively charged object A gets in contact with the neutral object B, the negative charge of object will induce the opposite charges on object B. Hence, there will be a positive charge on object B

8 0

3 years ago

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0

3 years ago

Derase

lisov135 [29]

Answer:

a. Power = 1000 Watts or 1 Kilowatts.

b. Current = 4 Amperes.

Explanation:

Given the following data;

Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules

Voltage = 250V

Time = 30 minutes to seconds = 30 * 60 = 1800 seconds

To find the power rating;

Power = energy/time

Substituting into the equation, we have;

Power = 1800000/1800

Power = 1000 Watts or 1 Kilowatts.

b. To find the current taken from the supply;

Power = current * voltage

1000 = current * 250

Current = 1000/250

Current = 4 Amperes.

8 0

3 years ago

An object in free fall near the surface of the earth accelerates at a rate of 78979.4 mi/hr What is the rate of acceleration for

Nina [5.8K]

Answer:

32.176 ft/s²

9.807 m/s²

Explanation:

Data provided in the question:

Acceleration of the free falling object = 78979.4 mi/hr²

a) In ft/s

Now,

1 mi = 5280 foot

and,

1 hour = 3600 seconds

thus,

78979.4 mi/hr = $\frac{\textup{78979.4}\times\textup{5280}}{\textup{1}\times\textup{3600}^2}$

78979.4 mi/hr = 32.176 ft/s² ...............(1)

b) In m/s²

Now,

1 foot = 0.3048 m

thus,

we have from 1

32.17 ft/s² = $\frac{\textup{32.17}\times\textup{0.3048}}{\textup{1}\times\textup{1}^2}$

78979.4 mi/hr = 32.176 ft/s² = 9.807 m/s²

7 0

3 years ago

Name the device of measurement and write its used or its function?​

Name the device of measurement and write its used or its function?