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Oksanka [162]
4 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]4 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.

You might be interested in
Akio draws the ray diagram shown.
Inessa [10]

Answer: Move the small car so it appears on the left side of the lens.

Explanation:

Because the lens is reflective the small car would apear on the same side as the normal car.

Hope this helps :)

3 0
3 years ago
A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the
tatiyna

Answer:

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁)................... Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t................... Equation 2

make t  the subject of the equation

t = Q/P................. Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.

6 0
3 years ago
Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?
Komok [63]
Data:

E = 20 V
R_{1} = 65\Omega
R_{2} = 65\Omega
R_{3} = 65\Omega
R_{4} = 65\Omega

<span>Now that we have all the values ​​we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

</span>R_{T} =  R_{1} + R_{2} + R_{3} + R_{4}
R_{T} = 65 + 65 + 65+ 65
R_{T} = 260\Omega

<span>Like this:
</span>
I_{T} =  \frac{E}{ R_{T} }

I_{T} = \frac{20}{ 260 }
I_{T} = 0,076923076...

\boxed{\boxed{I_{T} \approx 0,07A}} 
Answer:
<span>The intensity of the total electric current 
</span>\boxed{\boxed{I_{T} \approx 0,07A}} 

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.
4 0
3 years ago
A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble
Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

\vec{F_B} = q(\vec{V}\times \vec{B})

F_B = q|v||B| sin\theta

Here,

q = Charge

v = Velocity

B = Magnetic field

\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}

Our values are given as,

\theta = 35.7\°

q = 22.5*10^{-9}C

B = 1.05*10^{-5}T

v = 3.11m/s

Replacing,

F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)

F_B = 4.224*10^{-13}N

Therefore the size of the magnetic force acting on the bumble bee is 4.22*10^{-13}N

3 0
4 years ago
Question 8
viktelen [127]

Answer: D(t) = 8.e^{-0.4t}.cos(\frac{\pi }{6}.t )

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = a.sin(\omega.t) or y = a.cos(\omega.t)

where:

|a| is initil displacement

\frac{2.\pi}{\omega} is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

y=a.e^{-ct}.cos(\omega.t) or y=a.e^{-ct}.sin(\omega.t)

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

y=a.e^{-ct}.cos(\omega.t)

period = \frac{2.\pi}{\omega}

12 = \frac{2.\pi}{\omega}

ω = \frac{\pi}{6}

Replacing values:

D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)

The equation of displacement, D(t), of a spring with damping factor is D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t).

3 0
3 years ago
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