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2 years ago

Pls help me quick im doing a quiz

Physics

2 answers:

yawa3891 [41]2 years ago

7 0

Answer): A

I sorry if it is wrong

Katyanochek1 [597]2 years ago

6 0

ANSWER : 0.19 kilograms m/s south

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In the modern model of the atoms over 99.9% of the atoms is made up of:

QveST [7]

In the modern model of the atoms over 99.9℅ of the atom is made up of empty space

3 0

3 years ago

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Calculate the average speed between seconds 1 and 4 PLS HELP MEEEEEEEEE

sesenic [268]

Answer:

8.7

Explanation:

3 0

3 years ago

A certain gas occupies a volume of 3.7 L at a pressure of 0.91 atm and a temperature of 283 K. It is compressed adiabatically to

Nostrana [21]

Answer:

(a)

P₂ = 7.13 atm

(b)

T₂ = 157.14 K

Explanation:

(a)

V₁ = initial volume = 3.7 L = 3.7 x 10⁻³ m³

V₂ = final volume = 0.85 L = 0.85 x 10⁻³ m³

P₁ = Initial Pressure of the gas = 0.91 atm = 0.91 x 101325 = 92205.75 Pa

P₂ = Final Pressure of the gas = ?

Using the equation

$P_{1} V{_{1}}^{\gamma } = P_{2} V{_{2}}^{\gamma }$

$(92205.75) (3.7\times 10^{-3})^{1.4 } = P_{2} (0.85\times 10^{-3})^{1.4 }$

$P_{2}$ = 722860 Pa

$P_{2}$ = 7.13 atm

(b)

T₁ = initial temperature =283 K

T₂ = Final temperature = ?

using the equation

$P{_{1}}^{1-\gamma } T{_{1}}^{\gamma } = P{_{2}}^{1-\gamma } T{_{2}}^{\gamma }$

$(92205.75)^{1-1.4 } (283)^{1.4 } = (722860)^{1-1.4 } T{_{2}}^{1.4 }$

T₂ = 157.14 K

6 0

3 years ago

A "phaser" blast form the starship "ENTERPRISE" sends out a

Iteru [2.4K]

The frequency of the light beam is 4.72 x 10¹⁴ Hz.

<h3>Frequency of the light beam</h3>

The frequency of light beam and wavelength of light is related in the following equations;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency
λ is the wavelength

f = c/λ

f = (3 x 10⁸) / (636 x 10⁻⁹)

f = 4.72 x 10¹⁴ Hz

Thus, the frequency of the light beam is 4.72 x 10¹⁴ Hz.

Learn more about frequency of light beams here: brainly.com/question/14295752

5 0

2 years ago

You put a ball on the end of a string. You swing the ball

Sav [38]

Answer:

$19.6 ms^{-2}$

Explanation:

By the definition of centripetal acceleration ,

$a_c=\frac{V^{2} }{r}$

where ,

a = centripetal acceleration

V= velocity

r = radius of circle

$a_c=\frac{V^{2} }{r}[\\a_c=\frac{5.6^{2} }{1.6} \\=19.6 ms^{-2}$

3 0

3 years ago