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2 years ago

The physical property that determines that how easily heat and electricity pass through a material is?

Physics

1 answer:

Varvara68 [4.7K]2 years ago

6 0

Answer: Hello! I'm Jungkook. Here is your answer.....

A. conductivity

Explanation:

Hope this helps! Anneyong/Bye!

xoxoKookie

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What type of cooling would a scientist determine formed an igneous rock found with large crystals? Slow cooling Medium rate cool

Dmitriy789 [7]

Slower cooling engenders the growth of larger crystals in igneous rocks, thus, your answer should be slow cooling!

Hope this helped!

5 0

3 years ago

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If a soccer ball takes 20 seconds to roll 10 meters, what is the average speed of the soccer ball?

xxTIMURxx [149]

2 Seconds per 1 meter

5 0

3 years ago

If the mass of an object on earth is 40 kilograms it’s mass on the moon is

g100num [7]

Speaking bout mass,
it's mass is still 40Kg on the moon

but it's Weight on the moon is 1/6mg

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4 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

4 years ago

You illuminate a slit with a width of 71.7 μm with a light of wavelength 737 nm and observe the resulting diffraction pattern on

miskamm [114]

To solve this problem we will apply the concepts related to the double slit-experiment. Under this concept we understand the relationship between the minimum angle, depending on the order of the fringes, the wavelength and the distance between slits. Therefore we have the following relation,

$sin(\theta_{min}) = \frac{m\lambda}{D}$

Here,

m = Order of the fringes

D = Distance between slits

$\lambda$ = Wavelength

Replacing with our values we have,

$sin(\theta_{min}) = \frac{(1)(737*10^{-9})}{71.7*10^{-6}m}$

$sin(\theta_{min}) = 0.01028$

Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then

$Y=Lsin\theta$

$Y =2.27*0.01028$

$Y =0.0233m$

Thus the width of the central maximum is

$W=2Y=2*0.0233$

$W = 0.0466m$

Therefore the widht is 0.466m

6 0

4 years ago