All categories

2 years ago

What common laboratory measuring device will we likely always use in experiments that measure enthalpy changes?

Physics

1 answer:

adelina 88 [10]2 years ago

5 0

The most common measuring device to be used in measuring enthalpy changes is the thermometer.

You might be interested in

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

2 years ago

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t

arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector $n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

$\tau=4.7\times10^{-4}\ N-m$

Hence, The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

5 0

2 years ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0

2 years ago

Ingrid is moving a box from the ground into the back of a truck. She uses 20 N of force to move the box 5 meters. If she uses an

Oduvanchick [21]

Answer:

Explanation:

4 0

2 years ago

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

oksian1 [2.3K]

Answer:

$\phi _{B} =0.855 T-m^{-2}$

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

$B = \mu _{O}J$

magnetic flux is given as

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

2 years ago