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Fantom [35]
3 years ago
9

Consider three different resistors connected to a battery in parallel.

Physics
1 answer:
zhuklara [117]3 years ago
5 0

Answer:

c) All choices listed are definitely true.

Explanation:

a)

  • As they are connected in parallel, the voltage drop is equal in all the resistors, by definition of a parallel connection, and equal to the terminal voltage of the battery (assuming that the internal resistance of the battery is negligible).

b)

  • Applying Ohm's law to the circuit, we arrive to the following expression:

        I_{tot} = I_{1} +I_{2} + I_{3}  \\  I_{tot} =\frac{V}{Req} = \frac{V}{R_{1}} +\frac{V}{R_{2} }  +\frac{V}{R_{3}}

  • Simplifying common terms, we have:

       \frac{1}{Req} = \frac{1}{R_{1}} +\frac{1}{R_{2} }  +\frac{1}{R_{3}}

  • It can be seen that the equivalent resistance, is less than any of the resistances.

d)

  • Due to the charge conservation principle, as the current is made from moving charges, the sum of the currents passing through these resistors is equal to the current passing through the battery (KCL).
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An electric motor takes 30 s to lift a box of mass 45 000g to a height of 1.7 m. Calculate the power of the electric motor.
Sidana [21]

Power P is the amount of energy E transferred or converted per unit time t, and is expression is:

P = E/t

The necessary energy to lift the box its the work of against the Gravity Force, and is given by:

E = mgh

Then the power is:

P = E/t

P = mgh/t

P = (45 kg)(9.8 m/s²)(1.7 m )/ 30

P = 24.99 W

<h2>P ≈ 25 W</h2>
3 0
2 years ago
Which is an example of a multicellular organism? A. gorilla B. amoeba C. bacterium D. paramecium
Natali5045456 [20]
A gorrila is a multicellular organism
8 0
3 years ago
A dog runs 300m North and sees the dog catcher and runs 120 m south .Whats the dogs displacement.The same jogger runs 3 miles ar
ozzi

Answer:

<em>1. 180 m</em>

<em>2. 1.5 miles/hour</em>

<em>3. 0 m/s </em>

Explanation:

The complete question is

A dog runs 300 m North and sees the dog catcher and runs 120 m south Whats the dogs displacement.

A jogger runs north for 3.0 miles. If this took 2.0 hours, what is the jogger’s average velocity in miles per hour?

The same jogger runs 3.0 miles on the track, starting and finishing his run after completing exactly 12 laps around the track. What is his average velocity?

1. Displacement is how far the dog moves from the original starting point.

If the dog runs 300 m north, and then 120 m south, then the displacement from the starting point will be

displacement = 300 - 120 = <em>180 m</em>

2. Displacement of the jogger = 3 miles

Time taken by the jogger = 2 hours

velocity of the jogger = ?

Velocity is = displacement/time

velocity = 3/2 = <em>1.5 miles/hour</em>

3. velocity = displacement/time

if the jogger completes exactly 12 laps around the track, then the jogger will return to the starting point. This means that the jogger's displacement from the starting point is 0 miles

Jogger's velocity will therefore be = <em>0 m/s </em>

8 0
3 years ago
A pulley of diameter 49.0 in revolves at 1860 rpm. What is the diameter of a second pulley, in inches, if it rotates at 620 rpm?
Rudiy27

Answer;

First pulley;

Diameter; 49.0, revolutions 1860 rpm

Second pulley;

Diameter; x ; revolutions 620 rpm

To get the diameter of the second pulley;

= (Diameter of the first × Revolutions of the first)/ revolutions of the second.

= (49 ×1860)/620

= 147

Thus; the diameter of the second pulley is 147.

4 0
4 years ago
A 1 kg block of wood is attached to a spring, of force constant 200 N/m, which is attached to an immovable support. The block re
mixas84 [53]

Answer:

v=15.9499m/s

Explanation:

From the question we are told that:

Mass of wood m=1kg

force constants k= 200N-m

Coefficient of kinetic friction \mu= 0.2

Bullet mass m_b= 20 \approx 0.02kg

Spring compresion y=15cm \approx 0.15 m

Generally the equation for kinetic energy of bullet K>E_b is mathematically given by

Complete Question

K.E_b=spring potential energy+work done against friction

K.E_b=\frac{1}{2} mbv^2

\frac{1}{2} m_b v^2=\frac{1}{2} ky^2+\mu my

\frac{1}{2} (0.02)v^2=\frac{1}{2} (0.2)(0.15)^2+0.2(1)(0.15)

v=15.9499m/s

v\approx16m/s

5 0
3 years ago
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