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3 years ago

Which is an example of radiation? Check all

Physics

2 answers:

vampirchik [111]3 years ago

8 0

Answer:

Explanation:

Examples are;

Ultraviolet light from sun.

Heat from a stove burner.

X-ray from an x-ray machine.

Alpha particle emit from a radio active decay of uranium.

Sound waves from your stereo.

Microwave from micro oven.

ultraviolet light from a black light.

Gamma radiations from a supernova.

AND MANY MORE.

USPshnik [31]3 years ago

6 0

Answer:

The answer to the following question is actually A.

Explanation:

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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un

Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within 1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

$B_{i} =\frac{hr/2}{k}$

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

$\pi$ Dh(T-Tinf)= $I^{2} R_{e}$

make T the subject of the equation , we have

T=25+ $\frac{100^2*0.01}{\pi*0.001*500 }$

T=88.7 degC

rate of chnage in temperature

dT/dt= $\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)$

at t=o and integrating both sides $\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}$

we have

$\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}$

t=8.31s

steady state temperature =88.7deg C

t=time within 1 degC of it steady stae is 8.31s

7 0

3 years ago

How long was a 60 W light bulb turned on if it used a total of 580 J of energy?

icang [17]

Here's the tool you need. You can't answer the question without this:

   "1 watt"
means
   "1 joule of energy, generated, used, or moved, every second".

So 60 watts = 60 joules per second

   Total energy generated,
      used, or moved = (power) x (time).

   580 joules = (60 watts) x (time)

Divide each side
by (60 watts): Time = (580 joules) / (60 joules/sec)

   = (9 and 2/3) seconds .

7 0

3 years ago

What gas law looks at the relationship between pressure and temperature of a gas? A) Boyle's Law B) Charles' Law C) Gay Lussac's

Mandarinka [93]

Answer: it should be Boyle’s law

Explanation:

6 0

3 years ago

Read 2 more answers

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

6 0

3 years ago

. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe

julia-pushkina [17]

By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P₂*50

P₂*50= 300*75

P₂ = 300*75/50 = 450

P₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.

3 0

3 years ago

Read 2 more answers