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Oliga [24]
3 years ago
15

An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded t

he helicopter and flew to the corporate offices at an average speed of 75 mph. The entire distance was 156 mi. The entire trip took 3 h. Find the distance from the airport to the corporate offices.
Physics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

The distance from the airport to the corporate offices is 110 miles.

Explanation:

Let d be the distance from the airport to the office (at a velocity of 75 mph), the person travels an entire distance of 156 miles, 156-d is the distance from home to the airport (at a velocity of 30 mph).

The equaiton for velocity is:

Velocity=\frac{distance}{time}

From this you can find the equation for the time:

time=\frac{distance}{velocity}

You can use the total time to find the distance d because you know that the time between the home and the airport plus the time between the airport and the work is 3 hours:

3=\frac{156-d}{30}+\frac{d}{75}

You can solve the equation to find d:  

3=\frac{156*75-75d+30d}{30*75}

3=\frac{11700-45d}{2250}\\d=-\frac{(3*2250)-11700}{45}\\d=110

Now you have that the distance between the airport and the office is 110 miles.

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<em>Answer:</em>

Simple, there could be air in the package and volume would record that, whereas mass would count the mass of the cereal and discount the air.

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A quartz crystal vibrates with a frequency of 35,621 Hz. What is the period of the crystal's motion?
oksano4ka [1.4K]

Period = 1 / frequency

Period = 1 / (35,621 /s)

Period = 2.8073... x 10⁻⁵ sec

Period = <em>28.07 microseconds</em>

or

Period = <em>0.0281 millisecond</em>

4 0
3 years ago
Monochromatic light (vacuum = 612 nm) shines on a soap film (n = 1.33) that has air on either side of it. The light strikes the
weeeeeb [17]

Answer:

t  =   110.079 nm

Explanation:-

we know that

for constructive interference in thin film

2nt = (m + (1/2))λ

where, n is refractive index

for minimum thickness m =0

2nt = λ /2

thickness   t   = λ / 4n

Here  

wavelength of light, λ = 612*10^{-9} m

refractive indices = 1.393t   = \frac{(612*10^{-9})}{(4*1.33)}t   =   115.03*10^{-9} m

t  =   110.03 nm

4 0
3 years ago
025 10.0 points
mamaluj [8]

Answer:

m_b=278.73\ kg

Explanation:

<u>Center of Gravity </u>

It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below

\displaystyle x_c=\frac{\sum x_im_i}{\sum m_i}

Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at

\displaystyle x_c=\frac{(4.9)(90.4)+(2.45)(m_b)}{90.4+m_b}

\displaystyle x_c=\frac{442.96+2.45m_b}{90.4+m_b}

When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance  

x_b=1.2+2.45=3.65\ m

The man is located at  

x_m=1.2\ m

The center of gravity is computed now as

\displaystyle x_c=\frac{(1.2)(90.4)+(3.65)(m_b)}{90.4+m_b}

\displaystyle x_c=\frac{108.48+3.65m_b}{90.4+m_b}

Both centers of gravity are equal, thus

\displaystyle \frac{442.96+2.45m_b}{90.4+m_b}=\frac{108.48+3.65m_b}{90.4+m_b}

Simplifying

442.96+2.45m_b=108.48+3.65m_b

Rearranging

1.2m_b=334.48

Thus

\boxed{m_b=278.73\ kg}

6 0
3 years ago
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