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A mysterious object with a surface area of 0.015 m2, volume of 0.000125 m3, density of 100 kg/m3, specific heat of 100 J/(kgK), thermal conductivity of 2 W/(mK), with an unknown initial temperature was placed in a fluid with a density of 50 kg/m3, specific heat of 70 J/(kgK), thermal conductivity of 0.1 W/(mK), at a temperature of 400K. The heat transfer coefficient is given to be 10 W/(m2K). After 10 seconds, the temperature of the object is measured to be 380K. Determine the object's initial temperature.

Answers

Answer 1

Answer:

The object's initial temperature is 333.6 K

Explanation:

We first assume that the liquid can only transfer heat to the object through convective heat transfer method.

Let T₀ = the initial temperature of the object

T = temperature of the object at anytime.

The rate of heat transfer from the liquid to the object is given as

Q = -hA (T∞ - T)

T∞ = temperature of the fluid = 400 K

A = Surface area of the object in contact with the liquid = 0.015 m²

h = Convective heat transfer coefficient is given to be = 10 W/(m²K)

The rate of heat gained by the object is given by

mC (d/dt)(T∞ - T)

m = mass of the object = ρV

ρ = density of the object = 100 kg/m³

V = volume of the object = 0.000125 m³

m = ρV = 100 × 0.000125 = 0.0125 kg

C = specific heat capacity of the object = 100 J/(kgK)

The rate of heat loss by the liquid = rate of heat gain by the object

-hA (T∞ - T) = mC (d/dt)(T∞ - T)

(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)

- mC (dT/dt) = -hA (T∞ - T)

(dT/dt) = (hA/mC) (T∞ - T)

Let s = (hA/mC)

(dT/dt) = -s (T - T∞)

dT/(T - T∞) = -sdt

Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -st

(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ

(T - T∞) = (T₀ - T∞)e⁻ˢᵗ

s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12

T = 380 K at t = 10 s

T₀ = ?

T∞ = 400 K

st = 0.12 × 10 = 1.2

(380 - 400) = (T₀ - 400) e⁻¹•²

(-20/0.3012) = (T₀ - 400)

(T₀ - 400) = - 66.4

T₀ = 400 - 66.4 = 333.6 K

Hope this Helps!!!

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Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?

Answers

Answer:8pi

Explanatio

omega=2pi/T

Answer:

0000

Explanation:

How do lenses and mirrors compare in their interactions with light? A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.

Answers

This question involves the concepts of reflection and refraction.

The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".

LENSES AND MIRRORS

When it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.

Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.

Learn more about reflection and refraction here:

brainly.com/question/3764651

Answer:

C. lenses refract light; mirrors do not

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

Answers

Answer:

360 Nm

Explanation:

Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).

From the question,

The expression of torque is given as

τ = F×d......................... Equation 1

Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.

Given: F = 40 N, d = 9 m

Substitute into equation 1

τ = 40(9)

τ = 360 Nm.

Answer:

360Nm

Explanation:

Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;

τ = r F sin θ

Where;

r = distance from the pivot point to the point where the force is applied

F = magnitude of the force applied

θ = the angle between the force and the vector directed from the point of application to the pivot point.

From the question;

r = 9m

F = 40N

θ = 90° (since the force is applied perpendicular to the end of the bar)

Substitute these values into equation (i) as follows;

τ = 9 x 40 sin 90°

τ = 360Nm

Therefore the torque is 360Nm

The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is $6.4884 * 10^(-17) watt$

Calculation of the energy per second;

The area should be

$= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2$

Now

The power should be

$= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt$

Learn more about the energy here: brainly.com/question/14338287

Answer:

Power energy per second will be equal to $6.4884* 10^(-17)watt$

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing $I=1.2* 10^(-12)w/m^2$

Area is given by $A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2$

We know that power is given by $P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt$

So power energy per second will be equal to $6.4884* 10^(-17)watt$

If a freely falling object were somehow equipped with a speedometer, its speedreading would increase each second by
a) about 15 m/s.
b) a rate that depends on its initial speed.
c) a variable amount.
d) about 5 m/s.
e) about 10 m/s.

Answers

Answer:

e) about 10 m/s

Explanation:

Acceleration due to gravity is nominally* 9.8 m/s². That means the change in velocity each second is ...

(9.8 m/s²)(1 s) = 9.8 m/s ≈ 10 m/s

_____

* This is the value expected to be used in the solution of many math and physics problems. The standard value of 'g' on Earth is defined as 9.80665 m/s². It varies from place to place and with altitude. At any given place, it may also vary with time as a result of changes in mass distribution within the Earth.

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is: