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meriva
3 years ago
13

A vocalist with a bass voice can sing as low as 92 Hz.

Physics
1 answer:
Inessa05 [86]3 years ago
5 0

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

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the motion of a particle along a straight line is represented by the position versus time graph above. at which of the labeled p
atroni [7]

Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.

The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

Learn more about acceleration here: brainly.com/question/933224

Learn more: brainly.com/question/25887663

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2 years ago
If a 580 N net force acts on a 40 kg what will the acceleration of the car be
myrzilka [38]

Answer:

I think u have to do 580N times 40KG

7 0
2 years ago
Read 2 more answers
ok this question is for y'all how do what is the coriolis effect and what do it produce and how and i know the answer
Fudgin [204]
Idk can u explain to me?
6 0
3 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
What's 600,000,000 divided by 3,000.000,000,000?
Murljashka [212]

Answer:0.000002

Explanation: I Looked It Up lol

5 0
3 years ago
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