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NNADVOKAT [17]
3 years ago
5

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

. What angle, in radians, does the ladder make with the ground Round your answer to two decimal places
Physics
1 answer:
Papessa [141]3 years ago
7 0

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that sin(betha)=(leg opposite)/(hypotenuse)

In this case we will need to find β, this way:

betha=sin^-1((15ft/20ft))

Then β=48,6°  

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

betha=(48,6°)*(2pirads/360°)

then we find β=0,85rads

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Answer: Option C. Is the correct answer.

Explanation: Wave length is the distance between the crest of each wave. If you were to decrease it you would be decreasing the distance between each wave therefore making it smaller and closer together as option C. shows.

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3 years ago
A student using a tuning fork of frequency 512 Hz observes that the speed of sound is 340.0 m/s. What is the wavelength of this
Romashka [77]

Answer:

0.066m

Explanation:

Step one:

given

frequency =512 Hz

The speed of sound is 340.0 m/s.

Required

The wavelength

Step two:

the formula for wavelength is

v=f \lambda

\lambda= v/f

substitute the given data

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5 0
3 years ago
Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which
GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

8 0
4 years ago
The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between
gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

F = \frac{KQ_{1}Q_{2}}{d^{2}}

1.60 = \frac{4KQ^{2}}{d^{2}}    .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

F' = \frac{9KQ_{3}Q_{4}}{d^{2}}

F' = \frac{126KQ^{2}}{d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0
3 years ago
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