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seropon [69]
4 years ago
10

As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north

Physics
2 answers:
dedylja [7]4 years ago
7 0
<span>in this case velocity of airoplane will be 750*sin20'


</span>
Gwar [14]4 years ago
3 0

Answer:

v_y=71.25\ m/s

Explanation:

Given that,

Speed of an airplane, v = 750 km/hr = 208.33 m/s

An airplane is flying 20.0 degrees north of east, \theta=20^{\circ}

To find,

Velocity in north direction.

Solution,

Let v_y is the speed of the airplane to the north. The y component of any motion is given by :

v_y=v\ sin\theta

v_y=208.33\ m/s\ sin(20)

v_y=71.25\ m/s

So, the airplane is moving with a speed of 71.25 m/s to the north direction.

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What is the accletation of a 1500kg with a net force of 7500 N
matrenka [14]
Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.
3 0
3 years ago
An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre
andrezito [222]

Answer:

F=7.7\times10^2 N

Explanation:

The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A=1.10\times10^{-2} m^2

P=7\times10^{4} Pa.

Therefore, F=7\times10^{4}\times1.10\times10^{-2}.

F=7.7\times10^2 N

4 0
4 years ago
Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:    

1. Median

2. Lower quartile

3. Upper quartile  

4. Minimum value  

5. Maximum value  

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value  

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0
3 years ago
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egoroff_w [7]
If E = 1/2 * m * v^2
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6 0
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As light shines from air to water (n = 1.33) at an incident angle of 45.0 degrees what is the angle of refraction
erma4kov [3.2K]
Using Snell's law
we get
sin(I)/sin(r) = U2/U1
• where U2 represent the water's refractive index and U1 represent air's refractive index
thus
sin45°sin(r) = 1.33/1
1/√2*1.33 = sin(r)
1/1.88 = sin(r)
0.531 = sin(r)
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5 0
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