Answer:

**Answer:**

D is not the a vector quantities

Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

A vertical scale on a spring balance reads from 0 to 245 N . The scale has a length of 10.0 cm from the 0 to 245 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.55 Hz . Ignoring the mass of the spring, what is the mass m of the fish?

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine? Where, Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem W: the amount of work done by the engine during one cycle or during some time specified in the problem A) e=QhW B) e=QcQh C) e=QcW D) e=WQh E) e=WQc

A thin film soap bubble (n=1.35) is floating in air. If the thickness of the bubble wall is 300nm, which of the following wavelengths of visible light is strongly reflected?

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

A vertical scale on a spring balance reads from 0 to 245 N . The scale has a length of 10.0 cm from the 0 to 245 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.55 Hz . Ignoring the mass of the spring, what is the mass m of the fish?

In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine? Where, Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem W: the amount of work done by the engine during one cycle or during some time specified in the problem A) e=QhW B) e=QcQh C) e=QcW D) e=WQh E) e=WQc

A thin film soap bubble (n=1.35) is floating in air. If the thickness of the bubble wall is 300nm, which of the following wavelengths of visible light is strongly reflected?

**Answer:**

d' = 75.1 cm

**Explanation:**

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

or

**d' = 75.1 cm**

**So, the apparent depth is 75.1 cm. **

The apparent depth of a 1.00-meter-deep pool, when viewed from above, is around 75.2 centimeters. This difference is due to light **refraction** in water, causing optical distortion.

When observing a shallow pool of 1.00 meter depth from above, the apparent depth is altered by the phenomenon of light refraction in water. **Light** bends as it passes from air into water, affecting the way objects are perceived underwater.

The apparent depth is less than the actual depth due to this bending of light. To calculate the apparent depth, one can use the **Snell's** Law formula, which relates the angles of incidence and refraction to the refractive indices of the two media.

However, a simplified formula for the apparent **depth** (d') in terms of the actual depth (d) is given by d' = d/n, where 'n' is the refractive index of water (approximately 1.33). Therefore, in this case, the pool's apparent depth, when viewed from above, will be approximately 75.2 centimeters, making it shallower than it appears at first glance due to the optical effects caused by light traveling through water.

For more such questions on **refraction**

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Complete question below:

"What is the apparent depth, in centimeters, when looking straight down at a shallow pool that is 1.00 meter deep? Note that the apparent depth is different from the actual depth due to the refraction of light in water."

**Answer:**

4611.58 ft/s²

**Explanation:**

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

**Magnitude of acceleration while stopping is 4611.58 ft/s²**

**Explanation:**

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

A. Consult with a friend and get their feeback

B. Dispute the beliefs by asking if these are true and examining the evidence

C. Seek mental health counseling

D. It is just too hard so let's just forget it.

**Answer:**

i believe the answer is B

**Explanation:**

Seeking the right answer is the best thing to do

min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.40-

min interval.

**Answer:**

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

**Explanation:**

The wheel travels through

Θ = (7.40/37.3)*360º = 71.42º

and so the length of the line segment connecting the initial and final position is

L = 2*L*sin(Θ/2) = 2 * (183m/2) * sin(71.42º/2) = 107 m

so the average velocity is

v = L / t = 107m / 7.40*60s = 0.241 m/s

Initially, let's say the velocity is along the +x axis:

Vi = π * 183m / (37.3*60s) i = 0.257 m/s i

Later, it's rotated through 71.42º, so

Vf = 0.257m/s * (cos71.42º i + sin71.42º j) = [0.0819 i + 0.244 j] m/s

ΔV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] m/s = [-0.175 i + 0.244 j] m/s

which has magnitude

|ΔV| = √(0.175² + 0.244²) m/s = 0.300 m/s

Then the average acceleration is

a_avg = |ΔV| / t = 0.300m/s / (7.40*60s) = 6.76e-4 m/s²

The instantaneous acceleration is centripetal: a = ω²r

a = (2π rads / (37.3*60s)² * 183m/2 = 7.21e-4 m/s²

**Answer:**

**Explanation:**

Given that wheel completes one round in total time T = 37.3 min

so angular speed of the wheel is given as

now the angle turned by the wheel in time interval of t = 7.40 min

PART 1)

Now the average velocity is defined as the ratio of displacement and time

here displacement in given time interval is

R = radius = 91.5 m

Now time to turn the wheel is given as

now we have

PART 2)

Now average acceleration is defined as ratio of change in velocity in given time interval

here velocity of a point on its rim is given as

now change in velocity when wheel turned by the above mentioned angle is given as

time interval is given as

now average acceleration is given as

So lets fill out what we have first:

Vi or initial velocity = 20 m/s

Acceleration or a = 4 m/s^2

Time for the motion = 10s

Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:

Δx = Vi(t) + 0.5at²

Plug all of our information in:

Δx = (20)(10) + (0.5)(4)(100)

Δx = 400 m

**Answer:**

400 m

**Explanation:**

answer on ed