What is the efficiency of the machine in Figure 7.2 if the weight of the box is 200 N and required 98 N of effort Force?

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

$E=\frac{hc}{\lambda}$

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

6 0

3 years ago

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

Initial Condition

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

Boundary Conditions

At x=0

$T(0,t)=T_o$

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

At x=L

$-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

3 years ago

A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340

Fantom [35]

Answer:

The value is $P_G = 2.925 *10^{5} \ Pa$

Explanation:

From the question we are told that

The volume of the bottle is $v = 2 \ L = 2 * 10^{-3} \ m^3$

The gauge pressure of the air is $P_g = 340 \ kPa = 340 *10 ^{3} \ Pa$

Generally the volume of air before the bottle is turned upside down is

$V_a = \frac{3}{4} * V$

$V_a = \frac{3}{4} * 2 *10^{-3}$

$V_a = 0.0015 \ m^3 }$

Generally the volume air when the bottle is turned upside-down is

$V_u = \frac{5}{6} * 2 *10^{-3}$

$V_u = 0.00167 \ m^3$

From the the mathematical relation of adiabatic process we have that

$P_g * V_a^r = P_G * V_u^r$

Here r is a constant with a value r = 1.4

So

$340 *10 ^{3} * 0.0015^{1.4} = P_G * 0.00167^{1.4}$

$P_G = 2.925 *10^{5} \ Pa$

3 0

3 years ago

If a loaded truck that accelerates at 1 meter per second squared loses its load and has three-fourths of the origional mass, wha

Nina [5.8K]

I believe it'd accelerate at 1.25 m/s^2 instead of 1, as it lost 1/4 of its mass (.25), so now it is .25 of 1 faster.

8 0

3 years ago

two objects each weighing 2kg are lifted through the same distance in 4s and 1s respectively. which scenario requires more work?

Shalnov [3]

i believe it would be the same because time doesn't matter when calculating total work done.

Work=FxD

F=force

D=distance

7 0

3 years ago