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Alex17521 [72]
3 years ago
12

A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement to

the nearest tenth of a mile?
A.) -17.3 miles
B.) -10.0 miles
C.) 10.0 miles
D.) 17.3 miles

Physics
2 answers:
Virty [35]3 years ago
4 0
17.3 would be the correct answer.

Oduvanchick [21]3 years ago
4 0

Answer: The correct answer is option(D).

Explanation:

Distance travel by car in north of east = 20 miles

In a triangle ABC

Cos\theta =\frac{BC}{AC}

Cos 30^o=\frac{\sqrt{3}}{2}=\frac{BC}{20}

BC=17.32

The east component of the car’s displacement to the nearest tenth of a mile 17.32 miles.

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PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)
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3 0
2 years ago
PLEASE HELP:
german

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3.39 mins.

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7 0
2 years ago
The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of
kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\  g=9.8m/s^{2}

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m

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7 0
3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

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Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

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now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

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so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

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4 0
3 years ago
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