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horrorfan [7]
3 years ago
8

Will each of the following actions increase the distance between the diffraction spots? a) Increase the distance between the scr

een and the grating. (Yes or No) (1pt.) b) Rotate the grating by 90 degrees. (Yes or No) (1pt.) c) Use a different light source that has a longer wavelength. (Yes or No) (1pt.) d) Double the screen size. (Yes or No) (1pt.) e) Use a different grating that has a higher line density. (Yes or No) (1pt.)
Physics
1 answer:
diamong [38]3 years ago
5 0

Answer:

A) no

B) no

C) yes

D) no

E) yes

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Is cell using oxygen to break down sugar a physical or chemical change
Viefleur [7K]
Breaking down sugar (glucose) is a chemical change. Sugar is a compound that can be broken down.
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3 years ago
Read 2 more answers
A scientist shines light from a source onto a piece of metal, and no electrons are released by the metal. Increasing the intensi
Nezavi [6.7K]

This is the photoelectric effect, and it is best explained by the particle model of light.

<h3>What is the photoelectric effect?</h3>

The photoelectric effect refers to the emission of negatively charged particles and electromagnetic radiation that hits an object.

The photoelectric effect shows how electrons can be released from a given object when this material is absorbing electromagnetic radiation.

The photoelectric effect is a fundamental piece of evidence for understanding the nature of light particles.

Learn more about the photoelectric effect here:

brainly.com/question/1359033

7 0
2 years ago
P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0
3 years ago
If a 1 kg book has 46 Joules of gravitational potential energy how high is the shelf it is on?
Mashcka [7]

Answer:

4.7m

Explanation:

Given parameters:

Mass of the book  = 1kg

Gravitational potential energy  = 46J

Unknown:

Height of the shelf  = ?

Solution:

The potential energy is due to the position of a body above the ground.

        Gravitational potential energy  = mgh

m is the mass,

g is the acceleration due gravity  = 9.8m/s²

h is the height which is unknown

                       46  = 1 x 9.8 x h

                       h  = 4.7m

4 0
2 years ago
A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What
ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0
3 years ago
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