White light is sent through glass prism. Which color is refracted the most?.

HEY CAN ANYONE PLS ANSWER DIS!!!!!!

antiseptic1488 [7]

Answer:

I think That All the atoms of a given element are identical in that they have the same number of protons

Explanation:

3 0

3 years ago

Read 2 more answers

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

7 0

3 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

3 years ago

If my final exam is worth 30% and my class average is 80, what do I need to get to pass the exam?

kumpel [21]

You haven't told us what the passing percentage is on the exam,
or what the passing percentage is for the semester, or any of that.

5 0

3 years ago

What is 1113.28 in a sig fig?

Afina-wow [57]

Answer:

6 significant figure

Explanation:

The digits 111328 all are 6 figures with no figure being zero, neither zero after the other digits. In this case, all the numbers are significant and since they are only six numbers, then this is a six significant figure. In case we add another zero after digit 8, the zero is not significant but if added either infront of 8 or 2, the zero becomes significant.

4 0

3 years ago