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lakkis [162]
3 years ago
13

Household wiring often uses 2.0 mm diameter copper wires. The wires can get rather long as they snake through the walls from the

fuse box to the farthest corners of your house.
What is the potential difference across a 16 m long, 2.0 mm diameter copper wire carrying a 7.3 A current?
Physics
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

0.637 V

Explanation:

From Ohm's law,

V = IR ................ Equation 1

Where V = Voltage, I = current, R = resistance.

Also

R = Lρ/A........... Equation 2

Where L = Length of the copper wire, ρ = resistivity of the copper wire, A = cross sectional area of the copper wire.

But,

A = πd²/4.................. Equation 2

Where d = diameter.

Substitute equation 2 into equation 1

R = 4Lρ/πd²............... Equation 3

Substitute equation 3 into equation 1

V = I(4Lρ/πd²)........... Equation 4

Given: I = 7.3 A, L = 16 m, d = 2 mm = 0.002 m, π = 3.14,

constant: ρ = 1.72 x 10⁻⁸ Ωm

Substitute into equation 4

V = 7.3(4×16×1.72x10⁻⁸)/(3.14×0.002²)

V = 8.04×10⁻⁶/(1.256×10⁻⁵)

V = 0.637 V

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A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. (a) What is their v
nevsk [136]

Answer:

a) 16.86 m/s.

b) 15.40 m/s

c) 3.175 m/s  

Explanation:

let pi be the initial momentum of the system, pf be the final momentum of the system, m1 be the mass of the bullet and V1 be the intial velocity of the bullet, m2 be the mass of the block and V2 be the intial velocity of the block

a)  from the conservation of linear momentum:

                                          pi = pf

              m1×(V1)i + m2×(V2)i = Vf×(m1 + m2)

(0.0220)×(400) + (0.500)×(0) = Vf(0.0220 + 0.500)

                                         8.8 = 0.522×Vf

                                          Vf = 16.86 m/s

Therefore, the bullet-embedded block system will be moving with a speed of 16.86 m/s.

b)  let Vf be the final velocity of the bullet-embedded block system, Wf be the work done by friction on the system, Vi be the initial velocity that the bullet-embedded block system initially moves with.

the total work done on the system is given by:

                                Wtot = Δk = kf - ki

                                    Wf = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                    f×ΔxΔcos(Ф) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                 m×Δx×g×μ×(-1) =  1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                        m×g×Δx×μ =  1/2(Vi)^2 - 1/2(Vf)^2

1/2(0.0220 + 0.500)(Vf)^2 = 1/2(0.0220 + 0.500)(16.86)^2 - (0.0220 +           0.500)×(9.8)×(8)×(0.30)

                     (0.261)(Vf)^2 = 86.469

                                (Vf)^2 = 237.2196

                                       Vf = 15.40 m/s

Therefore,   bullet-embedded block system will have a speed of 15.40 m/s after sliding of the friction surface.

c) let Vf be the speed of the bullet-embedded before hitting the block, Vi be the initial velocity of the block and V be the velocity of the bullet-embedded block and block system.

                              M×Vf + m×Vi = (m + M)×V  

 (0.0200 + 0.50)×(15.40) + (2)×(0) = (0.022 + 0.50 + 2)×V

                                             8.008 = 2.522×V

                                                    V = 3.175 m/s  

  Therefore, the bullet-embedded block and block system will be moving at a speed of 3.175 m/s.

5 0
3 years ago
Which of the following would not be considered a projectile?
Naddika [18.5K]

Answer:

e. All of the above are projectile

Explanation:

A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.

5 0
3 years ago
What voltage would be measured across the 15 ohm resistor?
Alona [7]

Answer: That depends on what other components are in the same circuit

along with the 15-ohm resistor.

If there's nothing else, and the battery is connected across the 15-ohm

resistor, then when you measure the voltage across the 15-ohm resistor,

your voltmeter is also connected straight across the battery, and you read

10 volts.

Explanation:

6 0
3 years ago
If the net force acting on a moving object CAUSES NO CHANGE IN ITS VELOCITY, what happens to the object's momentum?
SOVA2 [1]

If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

<h3>What is momentum?</h3>

Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.

The momentum is the product of its mass and velocity.

This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.

Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

Learn more about momentum at: brainly.com/question/13554527

#SPJ1

5 0
2 years ago
Difference between effort and load​
scoray [572]

<u>Answer</u>:

Effort is the unaltered force. Load is the altered force.

4 0
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