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4 years ago

11

Two iron rods are placed horizontally, touching end-to-end. One end of one rod is heated. After some time, the other rod also fe

els warm to touch. How did the heat travel from the heated rod to the one that was not heated?
A) The air over the heated rod traveled to the other rod and made it warm.
B) Part of the heat was converted to radiation and traveled to the cooler rod.
C) Electrons within the rods started moving, distributing heat consistently in both rods.
D) The molecular movement in the first rod transferred energy to the molecules in the second.

2 answers:

algol134 years ago

6 0

The answer is D) <span>The molecular movement in the first rod transferred energy to the molecules in the second.</span>

saw5 [17]4 years ago

3 0

I can confirm that the answer is D

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Resistance and resistivity are related by a: 1.factor that is dependent on the length of the material 2.factor that is dependent

luda_lava [24]

Answer:

1. factor that is dependent on the length of the material.

2. factor that is dependent on the area of the material

Explanation:

Resistance R is directly proportional to its length L of an object and it is inversely proportional to the crossectional area of that object.

Mathematically, this is represented as:

a) Resistance R is directly proportional to its length L of an object

R ∝ L ......... Equation 1

b) Resistance R is inversely proportional to the crossectional area (A) of an object

R ∝ 1/A ............ Equation 2

Combining the two equations together, we have:

Resistance R is directly proportional to its length L of an object

R ∝ L/A .......... Equation 3

Resistivity of a material is the natural or inherent or intrinsic property that a material has. The Resistivity of a material is not dependent on the length or crossectional area of such material. It is only temperature dependent.

The resistance of a material is also dependent on the resistivity of the material. This means the resistance of a material is directly proportional to the resistivity of the material.

R ∝ ρ........Equation 4

Combining Equation 4 and 3 together, we have:

R = ρL/A

Where:

R = Resistance

ρ = Resistivity

L = Length of the material

A = Crossectional area of the material

Resistance and resistivity are related by a factor that is dependent on the length of the material a factor that is dependent on the area of the material

3 0

4 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

7. Delia Gonzalez, Mother of Felix, said Felix wanted to play Football but

Gelneren [198K]

Determinatio/Goal setting

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2 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

3 years ago

HELP ASAP!!!!! ILL MARK YOU BRAINLIEST!

Julli [10]

Hello! Sorry this is a little late!

The answer to your question would best be option C, y<span>es, because electric charges have electric fields surrounding them that allow them to exert forces on other objects without touching them.

I just took this test, and can 100% confirm this is the correct answer!

Hope this helps, and have a great day! :)</span>

8 0

3 years ago

Read 2 more answers