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sweet-ann [11.9K]
3 years ago
8

Planck’s constant is a ratio between which two quantities?

Physics
2 answers:
Stella [2.4K]3 years ago
8 0

Answer:

Energy and frequency

Explanation:

Readme [11.4K]3 years ago
7 0

Answer: D

Explanation:

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chubhunter [2.5K]
I would say it reflects the sun easily. That’s also how we see it :)
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3 years ago
How do i figure out this question?
nikklg [1K]

Answer:

0.75 g/cm^3

Explanation:

The formula for density:

\rho = \frac{m}{V}

Where m is the mass and V is the volume.

So, we can substitute values for m and V:

\rho = \frac{277}{370}\approx0.75

Therefore, the density is 0.75 g/cm^3 (watch the units!)

8 0
2 years ago
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Describe the relationship shown in each graph below
pochemuha

The distance decreases as the time increases

7 0
2 years ago
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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
2 years ago
How many significant figures does 0.00340 have?
True [87]
Greetings!

"<span>How many significant figures does 0.00340 have?"...

A significant figure is:
-A non zero number
-A zero in between non zero numbers
-Trailing zeros to the right of the decimal point

Therefore, this number has three significant figures:
3,4,0

Hope this helps.
-Benjamin</span>
4 0
3 years ago
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