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3 years ago

Does a solid sit by itself

Physics

2 answers:

Svetradugi [14.3K]3 years ago

8 0

Answer:

Yes?

Explanation:

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Gekata [30.6K]3 years ago

3 0

Answer: yes i think

Explanation:

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Sonar is a device that uses reflected sound waves to measure underwater depths. If a sonar signal has a frequency of 288 Hz and

mart [117]

Answer:

5.03 m

Explanation:

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency of the wave

For the sonar signal in this problem,

$f=288 Hz$

$v=1.45\cdot 10^3 m/s$

Substituting into the equation, we find the wavelength:

$\lambda=\frac{1.45\cdot 10^3 m/s}{288 Hz}=5.03 m$

3 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

What makes these image a represent of what research <br>

ycow [4]

The economic down fall

3 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher

AnnZ [28]

Answer:

Hello your question has some missing parts attached below is the complete question

answer : 4 μm

Explanation:

since the scale bar works the same way as a scale on a map , each bar will therefore represent 1 μm and the mature parent cell's is about 4 times the labeled value hence the Mature parent cell diameter will approximately be : 4 μm

8 0

3 years ago