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2 years ago

5

show how three identical 6 resistors must be connected tho have the following effective resistance values 9 and 4 ohms

1 answer:

SpyIntel [72]2 years ago

5 0

Answer:

connect two 9 ohms resistance in series now it becomes 18 ohm

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A car has a speed of 10 m/s and a mass of 1500 kg. what is the momentum of the car

Mrrafil [7]

Answer:

p = 15 newts

Explanation:

p is momentum

also there's a calculator to answer these questions.

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2 years ago

8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At

Mrac [35]

<u>Answer:</u>

Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

Velocity of car in first hour = 70 mph

Velocity of car in second hour = 60 mph

Velocity of car in third hour = 80 mph

Acceleration = Change in velocity / Time

Acceleration in second hour = (60 - 70)/1 = -10 mph²

Acceleration in third hour = (80 - 60)/1 = 20 mph²

So positive acceleration is in third hour and negative acceleration is in second hour.

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3 years ago

What does a velocity measurement include that a speed measurement does not? A: Time B:direction C:distance D: acceleration

andrew-mc [135]

Answer:

B. Direction

Explanation:

Speed is a scalar quantity and doesn't keep track of direction : Velocity is a vector quantity and is direction aware.

8 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is

Y_Kistochka [10]

Answer:

$0.11\times 10^{-6}weber$

Explanation:

We have given number of turns N = 560

Inductance L = 8.9 mH

Current through the coil = 7 mA

Inductance of the coil is given as $L=\frac{N\Phi }{I}$

Where N is number of turns I is current and $\Phi$ is flux

So $\Phi =\frac{LI}{N}=\frac{8.9\times 10^{-3}\times 7\times 10^{-3}}{560}=0.11\times 10^{-6}weber$

6 0

3 years ago

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