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LiRa [457]
3 years ago
11

A cylindrical container with a cross-sectional area of 85.2 cm 2 holds fluid having a density of 806 kg/m 3 . The surface of the

fluid is open to the atmosphere, and the pressure at the bottom of the tank is 106 kPa. What is the depth of the liquid?
Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

the answer would be B your welcome

Explanation:

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The temperature of the cosmic background radiation is measured to be 2.7 k. What is the wavelength of the peak in the spectral d
KATRIN_1 [288]

Answer:

1.07\cdot 10^{-3} m

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

\lambda=\frac{b}{T}

where

b=2.898\cdot 10^{-3} m\cdot K is Wien's displacement constant

T is the absolute temperature

For the cosmic background radiation, the temperature is

T = 2.7 K

So, the corresponding peak wavelength is

\lambda=\frac{2.898\cdot 10^{-3} m\cdot K}{2.7 K}=1.07\cdot 10^{-3} m

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3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
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3 years ago
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