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Dafna11 [192]
3 years ago
14

Which of the following would decrease the resistance to the flow of an electric current through a body?

Physics
2 answers:
seraphim [82]3 years ago
5 0

On page 37 it explains that If two conductors are composed of the same material and have the same cross-sectional area but different lengths, the electrical resistance of the longer conductor will be greater than that of the shorter conductor.

So the correct answer is A. shortening the conductor

Nookie1986 [14]3 years ago
3 0
R = ρ L/A. R= resistance, ρ= resistivity, L= length of the conductor. A = area of the conductor. Resistance is directly proportional to the length of the conductor. So if length of the conductor is decreased, resistance will also decrease. Hence A is the correct option
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The generation of a magnetic field by an electric current is _____
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The generation of a magnetic field by an electric current is _electromagnetism_.

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A falling raisin will have<br> whale. (more/less)<br> momentum than a falling blue
Natalka [10]

Answer:less

Explanation:thats the answer

3 0
3 years ago
A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later
ale4655 [162]

Answer:

140265.8 C = 1.403 × 10⁵ C

Explanation:

The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.

Potential work required to move the 757 kg car up a vertical height of 195 m = mgh

P.E = 757 × 9.8 × 195 = 1446627 J

Kinetic work done = (1/2)(m)(v²)

K.E = (1/2)(757)(25²) = 236562.5 J

Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J

And this would be equal to the potential of the battery.

For the battery, potential difference = (electric potential energy)/(charges moved)

ΔV = ΔU/q

q = ΔU/ΔV

ΔU = 1683189.5 J

ΔV = 12.0 V

q = 1683189.5/12 = 140265.8 C

7 0
3 years ago
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In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d
alex41 [277]

Answer:

v=18.89\frac{m}{s}

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)

Solving N from (2) and replacing in (1):

N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Replacing and solving for v:

gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}

4 0
3 years ago
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