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3 years ago

What is the SI unit for kinetic energy

Physics

2 answers:

OlgaM077 [116]3 years ago

5 0

The SI unit for kinetic energy is joules. Joules is measured in mass which is measured in kilograms.

trapecia [35]3 years ago

3 0

The easiest way to build a unit for energy is to remember that
'work' is energy, and
   Work = (force) x (distance).

So energy is    (unit of force) x (unit of distance)

   [Energy] =   (Newton) (meter) .

'Newton' itself is a combination of base units, so
energy is really
   (kilogram-meter/sec²) (meter)

   = kilogram-meter² / sec² .

That unit is so complicated that it's been given a special,
shorter name:
   Joule .

It doesn't matter what kind of energy you're talking about.
Kinetic, potential, nuclear, electromagnetic, food, chemical,
muscle, wind, solar, steam ... they all boil down to Joules.

And if you generate, use, transfer, or consume 1 Joule of
energy every second, then we say that the 'power' is '1 watt'.

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State four law of photoelectric effect

Bogdan [553]

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.

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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

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2 years ago

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A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

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3 years ago

Which elements do hydrogen fuel cells combine to produce electricity?

Fynjy0 [20]

Answer:

Hydrogen and Oxygen i believe

Explanation:

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3 years ago

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5. What's the temperature -10°C in kelvins?

konstantin123 [22]

Answer:

263.15

. . . . . . . . . .. ... . . . . . .. . . . . . . . ... .

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3 years ago

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba

lina2011 [118]

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$