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laiz [17]
4 years ago
8

A jet airliner moving initially at 503 mph

Physics
1 answer:
Varvara68 [4.7K]4 years ago
3 0

Answer:

The new speed is 1230.28 m/h

Explanation:

The jet airliner moving initially at 503 mph to the east

The wind is blowing at  855 mph in a direction 52° north of east

At first let us distribute the velocity of the wind into east component

and north component

→ The east component is 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Now we have two components of velocity in the east direction

and one component of velocity in the north direction

The new speed is the resultant of the east and north components

→ The east components are 503 m/h and 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Add the components of the speeds in direction of east

→ The east component = 503 + 855 cos(52) = 1029.39 m/h

→ The north component = 855 sin(52) = 673.75 m/h

Now we can find the new speed as a resultant speed of the east and

north components

→ The new speed = \sqrt{(1029.39)^{2}+(673.75)^{2}}

→ The new speed = 1230.28 m/h

<em>The new speed is 1230.28 m/h</em>

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Answer: 1.8\°

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d=3.4(10)^{-5}m is the width of the slit

\lambda=540 nm=540(10)^{-9}m is the wavelength of the light  

n is an integer different from zero.

Now, the second-order diffraction angle is given when n=2, hence equation (1) becomes:

dsin\theta_{2}=2\lambda (2)

Now we have to find the value of \theta_{2}:

sin\theta_{2}=\frac{2\lambda}{d} (3)

Then:

\theta_{2}=arcsin(\frac{2\lambda}{d})   (4)

\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})   (5)

Finally:

\theta_{2}=1.8\°   (6)

5 0
3 years ago
An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
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The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. an ostrich is run
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north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
<span>west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s</span>



6 0
4 years ago
suppose that you had an ohmmeter and a mystery component that could be either an inductor or a capacitor. How could you use the
sukhopar [10]

Answer:

To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is

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if the resistance = 0 then component is an inductor

Explanation:

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8 0
3 years ago
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