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Oksanka [162]
3 years ago
7

Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may displ

ay additional information when you click on them. Drag and drop each item into the correct column. Order does not matter.
ITEM BANK: Move to Bottom
A short, thick, cold wire is the best conductor.A shorter wire will allow electricity to move through at a higher rate than a longer wire.How well a material conducts current is an internal factor affecting resistance.If you double the length of a wire, you cut the resistance in half.If you double the thickness of a wire, you cut the resistance in half.Superconductors have no measurable resistance.The higher the temperature of the conductor, the lower the resistance.The resistance in a wire with less thickness is less.Thickness, length, and temperature are internal factors that affect resistance.When a light is first switched on, the light bulb's filament has a lower resistance than after it gives off light for awhile.
TRUE
FALSE
Physics
1 answer:
zavuch27 [327]3 years ago
7 0

Answer:

Explanation:

True                                                                              

How well a material conducts current is an internal factor affecting resistance.A shorter wire will allow electricity to move through at a higher rate than a longer wire.When a light is first switched on, the light bulb's filament has a lower resistance than after it gives off light for awhile.Superconductors have no measurable resistance.If you double the thickness of a wire, you cut the resistance in half.A short, thick, cold wire is the best conductor.

False

Thickness, length, and temperature are internal factors that affect resistance.The resistance in a wire with less thickness is less.The higher the temperature of the conductor, the lower the resistance.If you double the length of a wire, you cut the resistance in half.

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How much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate​
Nuetrik [128]

Answer:

<h3><em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>

<em>Sol</em><em>ution</em><em>,</em>

<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>

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<em>Now</em><em>,</em><em>.</em>

<em>weight = m \times g \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 100  \times 24.79 \\  \:  \:  \:  \:  \:  \:  = 2479 \: newton</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>

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5 0
3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is
Wewaii [24]

<u>Given </u><u>:</u><u>-</u>

  • An elevator is moving vertically up with an acceleration a.

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The force exerted on the floor by a passenger of mass m .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .

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For the FBD refer to the attachment . From that ,

\implies Weight_{apparent}= mg + ma

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2 years ago
If you are given distance and a period of time what can you calculate
Digiron [165]

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6 0
3 years ago
Read 2 more answers
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
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3 years ago
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