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3 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:

algol133 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]3 years ago

3 0

I can confirm that the answer is D

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Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.

Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

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2 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

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2 years ago

A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of

GREYUIT [131]

Given:

Sample 1:

Chloroform is $CHCl_{3}$

12 g Carbon

1.01 g Hydrogen

106.4 g Cl

Sample 2:

30.0 g of Carbon

Solution:

mass of chloroform from sample 1:

12 + 1.01 +106.4 =119.41 g

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mass of chloroform $\times$ $\frac{given mass of carbon}{mass of carbon atom}$

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2 years ago

Can A positively charged body attract another positively charged body

andriy [413]

Like charges repel, unlike charges attract

Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

So I would say no, unless the two bodies are placed close to each other where one has much more charge than the other, then due to induction, force of attraction becomes more than the force of repulsion.

3 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

2 years ago