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zmey [24]
3 years ago
13

This table shows the mass and volume of four different objects.

Physics
2 answers:
densk [106]3 years ago
7 0

X, Z, Y, W

ON EDG2020

suter [353]3 years ago
5 0

Answer:

Here its right but its also better than Barney's response

Explanation:

W, Y, Z, X or C

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A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000
Arlecino [84]

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

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2 years ago
Explain how the "natural frequency" of objects must be considered/analyzed in places like concert halls and airplanes.
hjlf
If you are talking about sound frequency you need to consider what area you are in because in a concert hall it is big and helps the sound spread but in an airplane it is the opposite. 
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5 0
3 years ago
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PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
A person drives a car around a circular road with a constant speed of 20 m/s. The
ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0
3 years ago
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