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Nady [450]
4 years ago
9

A plane leaves Seattle, flies 76.0 mi at 22.0 ∘ north of east, and then changes direction to 51.0 ∘ south of east. After flying

at 123 mi in this new direction, the pilot must make an emergency landing on a field. The Seattle airport facility dispatches a rescue crew.
In what direction should the crew fly to go directly to the field? Use components to solve this problem

How far should the crew fly to go directly to the field? Use components to solve this problem.
Physics
1 answer:
mote1985 [20]4 years ago
5 0
This can be solve by using a triangle, because the path of the plane formed a triangle. first solve the angle form by the second direction
angle = 180 - 51 - 22 = 107 degrees
then using the cosine law
c^2 = a^2 + b^2 - 2ab cos C
c^2 = 76^2 + 123^2 - 2 ( 76) ( 123) cos ( 107)
c = 162.4 mi <span>the crew fly to go directly to the field
</span>
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Motion equation of object is
Ganezh [65]

Take the derivative to find the velocity of the object:

v=\dfrac{\mathrm dx}{\mathrm dt}=-12+6t

The object stops when v=0:

-12+6t=0\implies6t=12\implies t=2

so the answer is E.

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A 1500 kg car moving with a speed of 20 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of
storchak [24]

Answer:

-100000 N.

Explanation:

Force: This can be defined as the product of the mass of a body and it's acceleration. The S.I unit of Force is Newton(N). The Formula of force is given as,

F = ma ........................... Equation 1

Where F = Average force exerted on the car, m = mass of the car, acceleration of the car, a = acceleration of the car.

a = (v-u)/t..................... Equation 2

Where v = Final velocity, u = Initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............. Equation 3

Given: m = 1500 kg, u = 20 m/s, v = 0 m/s (brought to rest), t = 0.3 s.

Substitute into equation 3

F = 1500(0-20)/0.3

F = 1500(-20)/0.3

F = -100000 N.

Note: The negative sign is due to the fact that the force exerted on the car by the pole is equal and opposite the force of the car.

7 0
3 years ago
In general, what do you think are the benefits if you already achieve your body goals.​
balandron [24]

Answer:

exercise dailyyy

Help you control your weight. ...

Reduce your risk of heart diseases. ...

Help your body manage blood sugar and insulin levels. ...

Help you quit smoking. ...

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Explanation:

7 0
3 years ago
(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg
valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

The weight of the buddy be m_2 = 120 kg

The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

∴

We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

4 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

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Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

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3 0
3 years ago
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