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Ilya [14]
3 years ago
11

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our

last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Physics
1 answer:
Grace [21]3 years ago
8 0

Answer:

The correct answer is - low pitch

Explanation:

Now for the case it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

if frequency increases then pitch will be increase as well as pitch depends on frequency.

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.

As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The
uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force F_{net} =\sum F = F_1 +F_2 +..... = ma

therefore we have from above equationF_{net} = F -f = ma

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

8 0
3 years ago
A 1.5 m3 tank contains 500 kg of liquid water in equilibrium with pure water vapor, which fills the remainder of the tank. The t
Stels [109]

Answer:

Explanation:

Attached are the solutions

7 0
4 years ago
El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una
Paraphin [41]

Answer:

 c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

                 

the area of ​​triangles is

           A = ½ base height

           A = ½ 4 3

          A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

           c² = b² + a²

           c = √ (4² + 3²)

           c = 5 m

this is the length of the curtain

5 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid
bixtya [17]

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

So

u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

6 0
4 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
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