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Mumz [18]
3 years ago
9

The note created by a flute will increase the speed of sound increases. When a marching band goes outside on a cold day, what wo

uld you predict would happen to the note played on a flute?
A) It would decrease because the speed of sound and temperature are proportional.
B) It would increase because the speed of sound and temperature are proportional.
C) It would decrease because the speed of sound and temperature are inversely proportional.
D) It would increase because the speed of sound and temperature are inversely proportional.
Physics
2 answers:
alexgriva [62]3 years ago
4 0

A).

It would decrease because the speed of sound and temperature are proportional.

Nutka1998 [239]3 years ago
3 0

Answer:

its A i just had this on usatestprep

Explanation:

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An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over
Lady bird [3.3K]

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

5 0
3 years ago
Read 2 more answers
1. Determine the energy released per kilogram of fuel used.
Serga [27]

The formula for energy release per kilogram of fuel burned is energy release per kg=6.702*10-13. and 19. J 1 Mev = 1.602 X 10 T

Calculate the energy in joules per kilogram of reactants given MeV per reaction. Energy is the ability or capacity to perform tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Think of a mole of plutonium-239 (molar mass: 239 grams) as a mole of "reactions."

Energy used in the US per person annually = 3-5 X 1011
Population (number of people) = 3.108The required mass of the fuel is 3.5x1011 x3-1x10 8x 10)/6.703 X1013 kg. the mass required: 1.62 x 1033 kg Mev in Joules 6 is equal to 101.60*I0-
19. J 1 Mev = 1.602 X 10 T, which translates to 1.602*1013/2.39x10-3 energy release per kilogram, or 6.702*10-13.

To learn more about Energy please visit -
brainly.com/question/27671072
#SPJ1

8 0
2 years ago
Which uses motion of water in nature to produce electricity?
Zinaida [17]

Answer: D. Hydroelectric power

Explanation: Hydroelectric means water power (not exactly but "hydro" is water and electric is well self explanatory)

6 0
3 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______
kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

5 0
3 years ago
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