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KonstantinChe [14]
3 years ago
11

How many 176 ohm resistors are connected in series to get a current of 5 a on 220 v battey

Physics
1 answer:
Phantasy [73]3 years ago
4 0
Total resistance = voltage / current = 220 / 5 = 44 ohms .

Even one 176-ohm resistor is too much.  The current through it is 1.25 A,
and more than one of them in series reduces the current even further.

Connecting them in <em><u>parallel</u></em>, however . . .

Four resistors of 176-ohms each, in <u>parallel</u>, have a net effective resistance
of 176/4 = 44 ohms ... exactly what you need to do the job.

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Three parachutists have the following masses: A: 50 kg, B: 40 kg, C: 75 kg Which one has the greatest terminal velocity?
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Answer:

A: 50 kg

Explanation:

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7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca
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Answer:

q = C V    charge on 1 capacitor

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A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

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