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Helen [10]
3 years ago
6

Three resistors, A, B, and C, are connected in parallel and attached to a battery, with the resistance of A being the smallest a

nd the resistance of C the greatest. Which resistor carries the highest current?
Physics
1 answer:
FromTheMoon [43]3 years ago
5 0

Answer:

Resistor A

Explanation:

Resistors A, B and C are connected in parallel to a battery of voltage represented by V.

Since they are in parallel, the same voltage, V, passes across the three of them.

And from Ohm's law, the voltage (V) through a resistor is the product of the current (I) flowing through it and the resistance (R) of the resistor. i.e

V = I x R

=> I = \frac{V}{R}

Therefore, assuming the values of the resistances of resistors A, B and C are A, B and C respectively,

(i) the current, I_{A} through A is

I_{A} = \frac{V}{A}

(ii) the current, I_{B} through B is

I_{B} = \frac{V}{B}

(iii) the current, I_{C} through C is

I_{C} = \frac{V}{C}

From the foregoing, it can be deduced that the current is inversely proportional to the resistance. Therefore, the higher the resistance, the lower the current. Consequential of this, the resistor that carries the highest current is the one with the smallest resistance, which is A

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Answer:

Time, t = 0.197 s

Solution:

As per the question:

Constant speed of the ball, v = 5.90 m/s

Coefficient of Kinetic friction, \mu_{k} = 0.105

Now,

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Friction force, f = \mu_{k}N

where

N = mg = normal reaction

Thus

f = \mu_{k}mg   (1)

Now, consider the ball to be a solid sphere,

Moment of inertia of the ball, I = \frac{2}{5}mR^{2}       (2)

where

R = radius of the ball

Torque is given by:

\tau = I\alpha          (3)

where

\alpha = angular\ acceleration

Thus from eqn (1), (2) and (3):

\alpha = \frac{5\mu_{k}g}{2R}

Now, the time taken is given by kinematic eqn:

\omega' = \omega + \alpha t

\omega' = 0 +\frac{5\mu_{k}g}{2R} t

\omega' = \frac{v}{r}

\frac{v}{r} = \frac{5\mu_{k}g}{2R} t

t = \frac{2\times 5.90}{5\times 1.05\times 9.8} = 0.197\ s

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4 years ago
Compare the events in the life of the Sun with those of a star that starts with less
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A star with greater mass will die out faster than the Sun.

<h3>What factors star is dependent on?</h3>

A star's future relies upon its mass. For the most part, the more huge the star, the quicker it consumes its fuel supply, and the more limited its life. The most huge stars can wear out and detonate in a cosmic explosion after two or three million years of combination.

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2 years ago
Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

\underset{V_{B}}{\rightarrow} = velocity of car B before collision = V_{B} i + 0 j

\underset{V_{AB}}{\rightarrow} = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

M_{A} = mass of car A = 1750 kg

M_{B} = mass of car B = 1450 kg

Using conservation of momentum

M_{A}  \underset{V_{A}}{\rightarrow} + M_{B}  \underset{V_{B}}{\rightarrow} = (M_{A} + M_{B}) ( \underset{V_{AB}}{\rightarrow} )

(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

V_{B} = 67.3 km/h        and  V_{A} = 34.4 km/



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Why is pitch an important property of music?
goldfiish [28.3K]
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3 years ago
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Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11
Nataliya [291]

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

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