All categories

3 years ago

Samples for a compound light microscope are typically prepared on _______.

Physics

1 answer:

elena55 [62]3 years ago

8 0

glass slides is the answer to your qestion

You might be interested in

(b) Which has more mass: 50 cm of gold or 50 cm of aluminum? Explain.

monitta

Gold’s molar mass is about 196 while aluminum is about 27, thus 50cm of gold has more mass

8 0

3 years ago

Use your data to predict what a 400g bag would weigh.

Kobotan [32]

3.92N

Explanation:

Given mass:

Mass of bag = 400g

The mass of a substance is the amount of matter it contains.

Weight of a substance is the vertical force acting on a body in the presence of gravity.

Weight = mass x gravity

Since the mass of the bag is 400g;

Now we convert to kg

1000g = 1kg

400g = $\frac{400}{1000}$ = 0.4kg

Weight of bag = 0.4kg x 9.8m/s²

Weight of bag = 3.92N

learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

6 0

3 years ago

Investigate the relationship between volume and pressure of a gas at a constant temperature.

sveta [45]

Gay Lussac's Law states: At a constant volume Pressure divided by Temperature isconstant P/T = k Together these three laws form the foundation of the Ideal GasLaw. Objective: Students will investigate Gay Lussac's Law relating pressure andtemperature at a constant temperature.

7 0

3 years ago

A jeweler is determining the optical properties of an unknown blue gemstone. She uses an angle of incidence of 62°, and measures

Juliette [100K]

Answer:

n = 1.76

Explanation:

According to the rule of ( n1 sin theta1 = n2 sin theta2 )

we know both angles so we insert them to the law and apply n1 = 1

so 1/2 = n2 sin 62 and we get the final answer

3 0

4 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago