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elena-14-01-66 [18.8K]
3 years ago
7

The number of wavelengths that pass a point each per second is:

Physics
1 answer:
Lesechka [4]3 years ago
3 0

Explanation:

d frequency I learned it last year

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A skier rides a ski-lift up to the top of the mountain, 1600 m above sea level. If the skier has a mass of 74 kg, how much poten
statuscvo [17]

Answer:

I think it is 1.16 MJ

Explanation:

PE=mgh

PE=(74)(9.81)(1600)= 1,161,504J = 1.16MJ

3 0
3 years ago
In which situation is the gravitational force between two objects hard to detect? (Options)
svlad2 [7]

A - the objects are too small

GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf

5 0
3 years ago
Read 2 more answers
What is the power of 10 when 0.00503 is written in scientific notation?
svetoff [14.1K]

Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

4 0
2 years ago
Read 2 more answers
Find the unit vector in the direction of vector B=4i+2j-5k​
solmaris [256]

Answer:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

Explanation:

for the unit vector, we need to divide the given vector by its norm, because it should be in the SAME direction as the original vector, but of magnitude "1".

We notice that the norm of the given vector is:

\sqrt{4^2+2^2+(-5)^2} =\sqrt{45}

Then, the unit vector becomes:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

7 0
2 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
3 years ago
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