PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity?
A. 926 m to the north
B. 5.2 m/s to the west
C. 46 m down
D. 12.3 m/s faster


Answer 1


D is not the a vector quantities

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The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be



d' = 75.1 cm


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,

n=(d)/(d')\n\nd'=(d)/(n)\n\nd'=(1\ m)/(1.33)\n\nd'=0.751\ m


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


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."

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 516 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.6 ft deep, estimate the magnitude of his average acceleration as he slowed to a stop (that is while he was impacting the snow).



4611.58 ft/s²


t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 32.174* 516+0^2)\n\Rightarrow v=182.218\ ft/s

v^2-u^2=2as\n\Rightarrow a=(v^2-u^2)/(2s)\n\Rightarrow a=(0^2-182.218^2)/(2* 3.6)\n\Rightarrow a=-4611.58\ ft/s^2

Magnitude of acceleration while stopping is 4611.58 ft/s²

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?



For Part (a)

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


=(v)/(c)\n v=(550-450)/(550)*3*10^(8)\n v=5.4545*10^(7)m/s

For Part (b)

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


=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s

So to deal with the irrational belief in REBT, we must Group of answer choices

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.



i believe the answer is B


Seeking the right answer is the best thing to do

The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.1. Find the magnitude of the average velocity at the wheel's rim, over a 7.40-
min interval.

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



Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²


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


v = 0.24 m/s

a = 6.75 * 10^(-4) m/s^2


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

so angular speed of the wheel is given as

\omega = (2\pi)/(T)

\omega = (2\pi)/(37.3) rad/min

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

\theta = \omega t

\theta = ((2\pi)/(37.3))(7.40) = 0.4\pi


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

here displacement in given time interval is

d = 2Rsin(\theta)/(2)

R = radius = 91.5 m

d = 183sin(0.2\pi) = 106.8 m

Now time to turn the wheel is given as

t = 7.40 min = 444 s

now we have

v = (d)/(t) = (106.8)/(444)

v = 0.24 m/s


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

v = R\omega

v = (91.5)((2\pi)/(37.3* 60))

v = 0.257 m/s

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

\Delta v = 2vsin(\theta)/(2)

\Delta v = 2(0.257)sin(0.2\pi)

\Delta v = 0.3 m/s

time interval is given as

t = 7.40 min = 444 s

now average acceleration is given as

a = (0.3)/(444)

a = 6.75 * 10^(-4) m/s^2

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.


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


400 m


answer on ed