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2 years ago

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belka [17]2 years ago

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Answer: marine current

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What is SI unit for distance

Dafna11 [192]

Answer:

meter

The SI unit of distance and displacement is the meter [m].

Explanation:

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3 years ago

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In the presence of ________, hydrogen peroxide will decompose into oxygen and water.

IrinaK [193]

Answer: Catalase enzyme

Explanation:

The catalase is an enzyme that is commonly found in the living organisms that are exposed to minute or high amount of oxygen or those in which aerobic respiration takes place.

This enzyme catalyzes the breakdown or decomposition of the hydrogen peroxide into oxygen and water. In humans and animals the catalase is present in liver, where it conducts the function of detoxifying the body by decomposing the toxins.

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3 years ago

What is the total displacement???

muminat

Displacement (between time 0 and time 25) is the area under the velocity time curve, i.e. ∫ vdt.
Here, v(0)=10, v(25)=34 (approx.)
Therefore
displacement = (1/2)(10+34 m/s)*(25-0) s [ trapezoid area ]
=550 m

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3 years ago

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Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A

kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × $\frac{ \lambda }{2}$ ........1

so λ = $\frac{2L}{10}$

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$ .................2

so

frequency for string A = f1 = $\frac{V1}{\lambda}$

f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

so

beat frequency is = f2 - f1

put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$ - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

beat frequency = 13.87 Hz

6 0

3 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

3 0

2 years ago