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Lilit [14]
3 years ago
11

A silver bar of length 30 cm and cross-sectional area 1.0 cm2 is used to transfer heat from a 100°C reservoir to a 0°C block of

ice.
How much ice is melted per second? (For silver, k = 427 J/s⋅m⋅°C. For ice, Lf = 334 000 J/kg.)
a. 4.2 g/s
b. 2.1 g/s
c. 0.80 g/s
d. 0.043 g/s
Physics
1 answer:
koban [17]3 years ago
4 0

Answer:

d. 0.043 g/s

Explanation:

Formula for rate of conduction of heat through a bar per unit time is as

follows

Q = k A ( t₁ - t₂ ) / L

A is cross sectional area and L is length of rod ,( t₁ - t₂ )  is temperature

difference . Q is heat conducted per unit time

Putting the values in the equation

Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²

= 14.23 J/s

mass of ice melted per second

= Q / Latent heat of ice

= 14.23 / 334000

= 0.043 g/s

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zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
2 years ago
You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen
Oksanka [162]

The de Broglie wavelength \lambda = 4.0\times 10^{-30}m

We know that

de Broglie wavelength = \lambda = \frac{h}{mv}\lambda = \frac{6.63\times 10^{-34}}{0.11\times 1.5 \times 10 ^{-3}}

\lambda = 4.0\times 10^{-30}m

<h3>What is de Broglie wavelength?</h3>

According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

Therefore, whether a particle is tiny or macroscopic, it will have a wavelength when examined.

The wave nature of matter can be seen or observed in the case of macroscopic objects.

To learn more about de Broglie wavelength with the given link

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3 0
1 year ago
explain why a high-speed collision between two cars would cause more damage than a low-speed collision between the same two cars
9966 [12]
The force equation can easily prove this.  F=ma.  This states that the force on an object is equal to mass times acceleration.  If the mass stays the same and the velocity of the cars increases than that means there is a larger force.  This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical.  Acceleration is the change in velocity over time.  If the velocity is higher with the same amount of time than that means there is a higher acceleration.  If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.
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8 0
3 years ago
Fifty points please help?
Zina [86]

it would be C laminated soda lime glass

6 0
2 years ago
If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?
butalik [34]

Answer:

The photon has a wavelength of 1.5x10^{-6}m

Explanation:

The speed of a wave can be defined as:

v = \nu \cdot \lambda (1)

Where v is the speed, \nu is the frequency and \lambda is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

c = \nu \cdot \lambda (2)              

 

Where c is the speed of light.    

Therefore, \lamba\lambda can be isolated from equation 2 to get the wavelength of the photon.

\lambda = \frac{c}{\nu} (3)

\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}

\lambda = 1.5x10^{-6}m

Hence, the photon has a wavelength of 1.5x10^{-6}m        

<em>Summary:  </em>

Photons are the particles that constitutes light.

3 0
3 years ago
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