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3 years ago

5

Newton's Law of Cooling says that the rate of cooling of an object is proportional to the difference between its own temperature

and the temperature of its surrounding. Write a differential equation that expresses Newton's Law of Cooling for this particular situation.

1 answer:

ASHA 777 [7]3 years ago

3 0

Answer:

$\frac{dQ}{dt} =-hA\Delta T(t)$

Explanation:Newton.s law of cooling states that the rate of cooling of an object is proportional to the difference between its own temperatures and temperature of its surroundings. Mathematically,

$\frac{dQ}{dt} =-hA [T(t)-T(s)]\\$

$\frac{dQ}{dt} =-hA\Delta T(t)$

where $Q$ is the heat transfer

$h$ is heat transfer coefficient

$A$ is the heat transfer surface area

$T$ is the temperature of the object's surface

$T(s)$ is the temperature of the surroundings

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5. An undisturbed soil sample has a wet density of 2.5 Mg/m3 when the water content is 25%. The specific gravity of the soil par

Dima020 [189]

Answer:

The submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³

Explanation:

Given;

wet density of soil sample = 2.5 Mg/m³ = 25 kN/m³

Specific gravity of solid particle = 2.7

The dry unit weight of soil;

$\gamma _d = \frac{\gamma _t}{1 +w} = \frac{25}{1+0.25} = 20 \ kN/m^3$

for undisturbed state, the volume of the soil is;

$V_s =\frac{\gamma _d}{G_s \gamma _w} = \frac{20}{2.7*9.81} = 0.76 \ m^3\\\\Void \ volume, V_v = 1-0.76 = 0.24 \ m^3$

$Void \ ratio, \ e = \frac{0.24}{0.76} = 0.32$

Submerged effective density is given as;

$\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e}$

density of water (ρw) = 2.7 x 25 kN/m³ = 67.5 kN/m³, substitute this in the above equation;

$\rho _b = \frac{\rho_w(\gamma_s -1)}{1+e} = \frac{67.5(2.7 -1)}{1+0.32} = 86.93 \ kN/m^3$

Therefore, the submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³

5 0

2 years ago

A construction worker dropped a brick from a high scaffolding. How fast was? a. How fast was the brick moving after 4.0 s of fal

Zigmanuir [339]

A) How fast was the brick moving after 4s?
Vf=?
Vi=0 (because it was dropped, not thrown)
A= -9.8m/s^2 (gravity)
t= 4s
Use the equation Vf=Vi+A(t)
Vf=0+(-9.8)(4)
Final answer: Vf= -39.2m/s
b) How far did the brick fall after 4s?
D=?
Vi=0
t=4s
A=-9.8m/s2
**You do have the final velocity, but it is best to avoid using numbers that you have calculated yourself.**
Use the equation: d=Vi(t)+0.5(A)(t)^2
d=(0)(4)+0.5(-9.8)(4)^2
d=(-4.9)(16)
d=-78.4m
Therefore, after 4s the brick fell 78.4m

5 0

3 years ago

If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x0?

tatiyna

ANSWER IS D X0=9.8m/s^2

5 0

1 year ago

A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t

WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

5 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago