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3 years ago

Describe the three methods of thermal energy transfer and provide a sketch that depicts each:

Physics

1 answer:

Jlenok [28]3 years ago

3 0

Answer:

a)the process by which heat or electricity is directly transmitted through the material of a substance when there is a difference of temperature

b)the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat

c)radiation is the emission or transmission of energy in the form of waves or particles through space or through a material medium.

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what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I

denis23 [38]

Answer: hello some part of your question is missing attached below is the missing detail

answer :

wf = M( v cos∅ )D / I

Explanation:

The Angular speed wf of the system after collision in terms of the system parameters and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) wf

where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )

In terms of system parameters and I

wf = M( v cos∅ )D / I

5 0

3 years ago

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic

olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$

$v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}$

$v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|$

$v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |$

$v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |$

$v = \frac{1}{5} | 250 - 50 + 15 |$

$v = \frac{215}{5}$

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0

3 years ago

Which of the following is a device designed to open an overloaded circuit and prevent overheating?

spin [16.1K]

Circuit breaker is a device that is designed to open an overloaded circuit and prevent overheating. The correct option in regards to the given question is option "a". The main purpose of the circuit breaker is to disconnect the defective switch from the circuit as soon as any problem arises. This helps in preventing any kind of major accident. Overheating can result in a big accident and the circuit breaker is the device that senses the overheating and disconnects the switch before any major incident. It is an automated system that acts as a protective devise.

4 0

3 years ago

Is Aluminium a metalliod?

svlad2 [7]

Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals, it has a close-packed crystalline structure and forms a cation in aqueous solution.

4 0

3 years ago

A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. how far will a point on the edge of the wheel

melisa1 [442]

First we find the angular acceleration $\alpha$ . The angular velocities are $\frac{240}{60} =4 \;rps$ and $\frac{360}{60} =6 \;rps$ The angular velocities and $\alpha$ are related as

$\omega=\omega_0+\alpha t\\ \alpha =\frac{\omega - \omega_0}{t} \\ \alpha =\frac{6-4}{(6.5)} \\ \alpha =0.3077 \;rps^2\\ \alpha =(0.3077 )2\pi =1.933 \;rad/s^2$

Angle turned in 6.5 seconds is

$2 \alpha\theta = \omega^2-\omega_0^2\\ \theta = \frac{ \omega^2-\omega_0^2}{2 \alpha}\\ \theta = \frac{ 6^2-4^2}{2 (1.933)}\\ \theta = 5.173 \; rad$

The distance traveled by a point on the edge of the wheel is $r \theta = \frac{33}{2}(5.173)= 85.35 \;cm$

7 0

3 years ago