Watchglass is a lab equipment that is used as a cover to prevent heated materials from splattering out of the container and as a holding plate for hot or flammable materials

<u>Explanation:</u>

Watch glass is an equipment used as a cover to prevent heated materials from splattering out of the container and as a holding plate for hot or flammable materials. It is a kind of concave glass which is also used to evaporate a liquid and also provides good air circulation which is used during cooking. The name watchglass was derived as they are see through and so similar to pocket glasses.

3 0

3 years ago

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

chubhunter [2.5K]

Answer:

The angular separation equals $0.35^{o}$

Explanation:

We have according to Snell's law

$n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})$

$\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})$

Using this equation for both the colors separately we have

$\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}$

Similarly for violet light we have

$\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}$

Thus the angular separation becomes

$\Delta \theta =14.48-14.13=0.35^{o}$

6 0

4 years ago

A cube of mass 128g has density of 2g/^3 what is the length each side of cube

Vilka [71]

The length of the side of the cube is 4 cm

Explanation:

The density of the cube is given by:

$\rho=\frac{m}{V}$

where

m is the mass of the cube

V is its volume

Also, the volume of the cube is given by

$V=L^3$ (2)

where L is the length of each side.

For the cube in this problem we have:

m = 128 g

$\rho=2 g/cm^3$

From the first equation, we find the volume of the cube:

$V=\frac{m}{\rho}=\frac{128}{2}=64 cm^3$

And now we use eq.(2) to find the length of the side:

$L=\sqrt[3]{V}=\sqrt[3]{64}=4 cm$

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4 0

4 years ago

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

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8 0

2 years ago