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melamori03 [73]

3 years ago

6

A 100 W lightbulb produces 5.0 W of visible light. (The other 95 W are dissipated as heat and infrared radiation.) What is the l

ight intensity on a wall 2.5 m away from the lightbulb?

2 answers:

Arada [10]3 years ago

6 0

17.5 would be your answer

lesantik [10]3 years ago

3 0

Answer:

17.5

Explanation:

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How is water vapor formed

Bas_tet [7]

Answer:

Y'now water vapor isn't half bad. It's like a vape but it can't affect your lungs. Anyway, Water vapor is water in gaseous instead of liquid form. It can be formed either through a process of evaporation or sublimation.

4 0

3 years ago

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A hot-water stream at 80 ℃ enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold wat

rewona [7]

Answer:

$\dot{m_{2}}=0.865 kg/s$

Explanation:

$\dot{m_1}= 0.5kg/s$

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

$\dot{m_{in}}=\dot{m_{out}}$

$\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}$

according to energy balance equation

$\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}$

$\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s$

4 0

4 years ago

Why does a rock weigh less in water than it does on land?

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Water is more dense then air so it sorta holds the rock as it sinks

7 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap

eduard

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

4 0

4 years ago

Read 2 more answers