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vodka [1.7K]
4 years ago
9

Karen has a mass of 51.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis

tance of 63.1 m, the longest escalator in the free world. The acceleration of gravity is 9.8 m/s 2 . How much work did the escalator do on Karen if it has an inclination of 26◦ ? Answer in units of J.
Physics
1 answer:
kifflom [539]4 years ago
4 0

Answer:

28852 J

Explanation:

When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.

The work (W) is:

W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.

Knowing that cos(26°) = 0.899, and F = m*g

W = 51.9*9.8*63.1*0.899

W = 28852 J

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Brilliant_brown [7]
The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
P=2\pi.R
Know recall that:
v=\frac{d}{t}
We now need to know how much distance is traveled during a minute or 60 seconds:
D=7200\times 2\pi\times R
Finally we divide this result with t=60 seconds:
v=\frac{7200\times2\pi\times R}{60}
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4 years ago
2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,
Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

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3 years ago
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3 years ago
A person who is good at caring for others might enjoy a career as a
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3 years ago
How to calculate depth of sea whan a ship receives sound produced by its fathometer after 3.5 second
HACTEHA [7]

Answer:

2600 m

Explanation:

A fathometer produces a sound wave and then detects the echo.  It takes 3.5 seconds for the echo to reach the ship, so that means it takes half the time (1.75 seconds) to reach the ocean floor.

The speed of sound in seawater is approximately 1500 m/s, so the depth of the ocean at that point is:

d = 1500 m/s × 1.75 s

d = 2625 m

Rounding to two significant figures, the depth is approximately 2600 m.

5 0
4 years ago
Read 2 more answers
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