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netineya [11]
3 years ago
10

Which device provides electrical energy to run an electric circuit

Physics
2 answers:
REY [17]3 years ago
8 0

The correct answer is

C. The battery

The battery is a device that provides a potential difference in the circuit, and so an electromotive force (e.m.f.) which pushes the electrons in the circuit from the negative pole towards the positive pole of the battery, so they move through the circuit. Therefore, it provides electrical energy.

Vinil7 [7]3 years ago
6 0
The energy comes from the battery, power supply, generator, wall outlet, solar panel, or windmill.
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A bag of sugar has a mass of 2.2kg what is its weight on earth
NARA [144]
The earth's gravitational acceleration is 9.8 m/s^2, and force = mass * acceleration. Therefore, a bag of sugar weighing 2.2 kg on earth would weigh 2.2 * 9.8 = 21.56 Newtons
8 0
3 years ago
Read 2 more answers
Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.
Natalka [10]

Explanation:

Things you need to know:

Accuracy refers to the maximum error encountered when a particular observation is made.

Error in measurement is normally one-half the magnitude of the smallest scale reading.

Because one has to align one end of the rule or device to the starting point of the measurement, the appropriate error is thus twice that of the smallest scale reading.

Error is usually expressed in at most 1 or 2 significant figures.

Tape

Equipment: It is made up of a long flexible tape and can measure objects or places up to 10 – 50 m in length. It has markings similar to that of the rigid rule. The smallest marking could be as small as 0.1 cm or could be as large as 0.5 cm or even 1 cm.

How to use: The zero-mark of the measuring tape is first aligned flat to one end of the object and the tape is stretched taut to the other end, the reading is taken where the other end of the object meets the tape.

Ruler

Equipment: It is made up of a long rigid piece of wood or steel and can measure objects up to 100 cm in length. The smallest marking is usually 0.1 cm.

How to use: The zero-end of the rule is first aligned flat with one end of the object and the reading is taken where the other end of the object meets the rule.

Vernier Caliper

Equipment: It is made up of a main scale and a vernier scale and can usually measure objects up to 15 cm in length. The smallest marking is usually 0.1 cm on the main scale.

It has:

a pair of external jaws to measure external diameters

a pair of internal jaws to measure internal diameters

a long rod to measure depths

How to use: The jaws are first closed to find any zero errors. The jaws are then opened to fit the object firmly and the reading is then taken.

Micrometer Screw Gauge

Equipment: It is made up of a main scale and a thimble scale and can measure objects up to 5 cm in length. The smallest marking is usually 1 mm on the main scale (sleeve) and 0.01 mm on the thimble scale (thimble). The thimble has a total of 50 markings representing 0.50 mm.

It has:

an anvil and a spindle to hold the object

a ratchet on the thimble for accurate tightening (prevent over-tightening)

How to use: The spindle is first closed on the anvil to find any zero errors ( use the ratchet for careful tightening). The spindle is then opened to fit the object firmly (use the ratchet for careful tightening) and the reading is then taken.

5 0
3 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
3 years ago
A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper
wolverine [178]
The resistance of the lamp is apparently  50V/2A  =  25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be      200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms 
in series with the lamp.  Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates  ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it. 
It has to be a special 'power resistor'. 
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms.  Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
3 0
3 years ago
How to change V=72km/hr to m/s
lapo4ka [179]
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr.  They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

=  20 meter/second
7 0
3 years ago
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