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lina2011 [118]
3 years ago
10

If the energy in a mechanical wave increases or decreases, what is also going to increase or decrease? A wavelength B amplitude

с frequency d refraction​
Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

i think wave lengh not 100% sure tho

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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c
seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2

So the volume flow rate along the pipe is

\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s

We can use the similar logic to find the cross-section area at the refinery

A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2

The radius of the pipe at the refinery is:

A_r = \pi r^2

r^2 =A_r/\pi = 0.446/\pi = 0.141

r = \sqrt{0.141} = 0.377m

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0
3 years ago
What is the primary visible color of an emission nebula?
zzz [600]
In emission nebulae, there are interstellar clouds of hydrogen, which glow red because of the intense radiation of hot stars inside the nebula
7 0
4 years ago
A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge
Troyanec [42]

Answer:

I_{l} =44.84 \mu A

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire =  1.30×10⁻² Ω  

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))

\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{V}{R}

I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}

I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))

I_{l} =44.84 \times 10^{-6}\A

I_{l} =44.84 \mu A

3 0
3 years ago
The radius of a conducting wire is doubled .What will be the ratio of its new specific resistance to the old one?
sladkih [1.3K]
The equation for the resistance R is: R=ρ*(l/A), where, ρ is electrical resistivity, l is the length of the conductor, and A is the surface area.

The initial surface area is:

A=r²π, then when we double the radius we get:
 
A₁=(2*r)²π=4*r²π=4*A

Initial resistance is: R=ρ*(l/A).

When we double the radius, resistance is: R₁=ρ*{ l / (4*A) }

The ratio of the new resistance to the old one:

R₁/R=[ρ*(l/A)] / [ ρ* { l / (4*A) } ] = ρ, l and A cancel out and we get:

R₁/R=(1/1)/(1/4)=4/1 
7 0
3 years ago
Alpha Particles
UNO [17]

Alpha > Beta > Gamma

Explanation:

Alpha particles are the heaviest radiation whereas gamma rays are the lightest.

An alpha particle closely resembles a helium particle.

Beta particles are similar to electrons in nature

Gamma rays have no weight. They are just rays.

  Alpha particles have a mass of  4.0012amu which is that of helium

  Beta particles have a mass of 5.5 x 10⁻⁴ amu

   Gamma rays have no mass.

Learn more:

Electromagnetic radiations brainly.com/question/6818046

#learnwithBrainly

7 0
3 years ago
Read 2 more answers
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