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Shkiper50 [21]
3 years ago
8

A package of chips weighs 45 grams. How much is that in kilograms?

Physics
2 answers:
stealth61 [152]3 years ago
7 0
0.045 kilograms that's the answer conversion is always so easy because you can always look it up
nirvana33 [79]3 years ago
5 0
45 grams is 0.045 kilograms
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What is true about energy that is added to a closed system
arsen [322]
It does work or increases thermal energy
4 0
3 years ago
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If 30ml of a fluid has a mass of 63g what is it’s density
murzikaleks [220]

Answer:

The answer to your question is 2.1 g/ml

Explanation:

Data

volume = 30 ml

mass = 63 g

density = ?

Process

Density is defined as the mass per unit volume. The units of density are g/ml or kg/m³.

Formula

Density = mass / volume

Substitution

Density = 63 / 30

Result

Density = 2.1 g/ml

7 0
3 years ago
What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he
andreyandreev [35.5K]

The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem!

<u>Given:</u>

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

<u>Unknown:</u>

fo = ?

<u>Solution:</u>

\Phi = h \times fo

3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo

fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )

\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544
  • Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

8 0
3 years ago
Read 2 more answers
Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find
True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0
3 years ago
A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as he
frutty [35]

Answer:

n = 1.89 x 10¹⁹ photons/s

Explanation:

given,

Power of bulb = 100 W

Visible light = 5 W

wavelength of the visible light = 600 nm

number of photons emitted per second

using formula of wavelength

\lambda = \dfrac{c}{\nu}

\nu= \dfrac{c}{\lambda}

\nu= \dfrac{3 \times 10^8}{600 \times 10^{-9}}

\nu=5 \times 10^{14}\ Hz

energy of photon

U = h υ

U = 6.63 x 10⁻³⁴ x 4 x 10¹⁴

U = 2.65 x 10⁻¹⁹ J

number of photon

n = \dfrac{P}{U}

n = \dfrac{5}{2.65 \times 10^{-19}}

n = 1.89 x 10¹⁹ photons/s

6 0
3 years ago
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