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FinnZ [79.3K]
3 years ago
8

Which of these objects are malleable? Check all that apply.

Physics
2 answers:
g100num [7]3 years ago
6 0

Answer:

The tin fork and knife, the copper coin, and the steel fence pole.

Explanation:

Those are both what people would call soft metals so they are malleable to the extent of probably not needing heavy duty equipment. It depends on you description of malleable because the steel fence pole could be malleable with the correct equipment and not snap in half if bent slowly enough.

The definition of malleable: (of a metal or other material) able to be hammered or pressed permanently out of shape without breaking or cracking.

But the glass table, marble sculpture and antique ceramic vase are nowhere near malleable because if you tried bending them they wouldn't bend but would shatter and break into pieces.

Anna35 [415]3 years ago
3 0

Answer:

1,3,6

Explanation:

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I think it’s A, I’m so sorry if I’m wrong.
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3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr
mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

f' = (\frac{v}{v\pm v_s})f

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

v_s is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

v_s = -40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz

When the train has passed the platform, we have

v_s = +40.4 m/s

So the frequency heard by the observer on the platform is

f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz

Therefore the overall shift in frequency is

\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

v=\lambda f

where

v is the speed of the wave

\lambda is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m

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When a helicopter engine spins the main rotor, it generates torque (see How a Helicopter Works), an equal and opposite reaction. Torque makes it so that the engine itself wants to spin.
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Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the sun.
pogonyaev
Total mass of the Sun = 2x10^30kg 

<span>So 76% of that = (2x10^30kg)*(0.76) = 1.52x10^30kg ----> total amount of Hydrogen i</span><span>f only 12% of that is used for fusion, then (1.52x10^30kg)*(0.12) = 1.82x10^9kg</span>
8 0
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