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Alenkinab [10]
3 years ago
5

As food passes through the alimentary canal, the presence of chemical secretions will occur in which order?

Physics
2 answers:
Natalija [7]3 years ago
7 0
The first one....................
Gekata [30.6K]3 years ago
5 0

It's B I took this test on FLVS

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Units called BEATS measure the loudness of sounds.<br> true or false
gogolik [260]

Answer:

False.

Explanation:

Decibels (dB) measure sound levels

7 0
3 years ago
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Answer:

3rd picture straight line going up right

Explanation:

3rd picture

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3 years ago
The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d
d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0
3 years ago
How are momentum and collision related?
Studentka2010 [4]

Answer:

im pretty sure a collision is a transfer of momentum.

Explanation:

hope this helps :)

5 0
3 years ago
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
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