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lions [1.4K]
3 years ago
11

You want to estimate the height of the Empire State Building. You start at its base and walk 15 m away. Then you approximate the

angle from the ground at that point to the top to be 88 degrees. How tall do you estimate the Empire State Bulding to be?
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

Answer:

429 m

Explanation:

In this exercise you are suppose to simulate a right triangle, the catheti are the distance you walked and the height of the empire state building.

tan(88) = (height) /(distance you walked)

28.64 = height / 15

height = 429 m

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How long does water evaporation take? What factors influence it?
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There is no certain time on how long it takes. Because the factors will always be different and the factors heavily affect the evaporation time. Some factors include: humidity, heat, how the sun is visible (whether clouds are covering it or not)
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A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050
Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

R=0.05Ω/Km*40km

R=2Ω

P=1000 MW

a).

P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k }  =8696.65A

Using law ohm

b).

V=I*R\\I=\frac{V}{R}

P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6}  W

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4 years ago
3. The expression 0.62 x10^3 is equivalent to...
Korolek [52]

\\ \sf\longmapsto 0.62\times 10^3

\\ \sf\longmapsto 62\times 10^{-2}\times 10^3

\\ \sf\longmapsto 62\times 10^{-2+3}

\\ \sf\longmapsto 62\times 10^1

\\ \sf\longmapsto 62\times 10

\\ \sf\longmapsto 620

5 0
3 years ago
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A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a
Snezhnost [94]

Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t  eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5  eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6  eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

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9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

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