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Dmitriy789 [7]
3 years ago
13

Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b

all misses the building on the way down, how long will it take from you threw the ball until it lands on the ground below? Give your answer in seconds and round the answer to three significant figures.
Physics
1 answer:
RSB [31]3 years ago
4 0
First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula 
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds
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Here,

m = Number of order bright fringe

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Both distance are the same, then

y_1 = y_2

\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}

\frac{m_1\lambda_1}{m_2\lambda_2} =1

 Rearranging to find the second wavelength

m_1 \lambda_1 = m_2 \lambda _2

\lambda_2 = \frac{m_1\lambda_1}{m_2}

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2 years ago
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3 years ago
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2 years ago
Tenemos un Cable de cobre de 1 km de longitud cuya sección es de 2 milímetros al cuadrado y queremos saber la resistencia que se
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Answer:

8.5 Ω

Explanation:

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