A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, concentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?A) a darkspotB) a bright spot thatis some color other than redC) a bright redspotD) a rainbow of color

Answers

Answer 1

Answer:

Answer:

After passing through the glass plate, the red light disperses and meets at point.

The convex lens has two refracting surfaces, and convex kens is called as converging lens. So, at the exact center of the lens, one observes a Dark spot.

Thus, the correct option is a) one observes a dark spot.

Answer 2

Answer:

Answer:

The answer is: A) a darkspot

Explanation:

When the red light passes through the glass plate, it is scattered. the convex lens (convergent lens) has two refractive surfaces, therefore, in the center of the lens, a characteristic dark spot would be observed.

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What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point45 N450 N450 kg10N
g A particular guitar string has a length of 60.0 cm, and a mass per unit length of 2.00 grams/meter. You hear a pure tone of 660 Hz when a particular standing wave, represented by the animation above, is excited on the string. Calculate the wavelength of this standing wave.
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Answers

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Final answer:

Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

Explanation:

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

Positive terms:

ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.

Negative terms:

ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.

Zero terms:

ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

Learn more about Energy equation here:

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

A 1300-turn coil of wire that is 2.10 cm in diameter is in a magnetic field that drops from 0.130 T to 0 T in 12.0 ms. The axis of the coil is parallel to the field.Question: What is the emf of the coil? (in V)

Answers

Answer:

4.875 V

Explanation:

N = 1300

diameter = 2.10 cm

radius = half of diameter = 1.05 cm

B1 = 0.130 T

B2 = 0 T

t = 12 ms

According to the law of electromagnetic induction,

$e = - N(d\phi )/(dt)$

Where, Ф be the magnetic flux linked with the coil

$e = - NA (dB )/(dt)$

$e = -1300*3.14*{1.05* 1.05* 10^(-4)*(0-0.130)/(12*10^(-3))=$

e = 4.875 V

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:
Find the radius and period of the orbit.

Answers

Answer:

r = 2,026 10⁹ m and T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

F = m a

where the force is electric

F = $k (q_1q_2)/(r^2)$

Acceleration is centripetal

a = v² / r

we substitute

$k (q_1q_2)/(r^2) = m (v^2)/(r)$

r = $k (q_1q_2)/(m \ v^2)$ (1)

let's look for the charge in the insulating sphere

ρ = q₂ / V

q₂ = ρ V

the volume of the sphere is

v = 4/3 π r³

we substitute

q₂ = ρ $(4)/(3)$ π r³

q₂ = 3 10⁻⁹ $(4)/(3)$ π 4³

q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)

r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

v = x / t

the distance traveled in a circle is

x = 2π r

In this case, time is the period

v = 2π r /T

T = 2π r /v

let's calculate

T = 2π 2,026 10⁹/628 103

T = 2.027 10⁴ s

Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.

Answers

Explanation:

The classic model of a black body made predictions of the emission at small wavelengths in open contradiction with what was observed experimentally, this led Planck to develop a heuristic model. This assumption allowed Planck to develop a formula for the entire spectrum of radiation emitted by a black body, which matched the data.

A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but the plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

Answers

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

$(V_1)/(T_1)=(V_2)/(T_2)\n\Rightarrow {V_2}=(V_1)/(T_1)* T_2\n\Rightarrow {V_2}=(12)/(298.15)* 373.15\n\Rightarrow {V_2}=15.01 L$

∴ Final volume at 100°C is 15.01 Liters.

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken