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3 years ago

What is the speed of a helicopter that traveled 1200 miles in 7 hours

Physics

2 answers:

zlopas [31]3 years ago

8 0

1200
-------=171 miles per hour
7

sladkih [1.3K]3 years ago

3 0

Answer:

The answer is 171 m/h

Explanation:

Speed relates to positional change with time, that is, it expresses the distance traveled in a given time.

The speed is expressed as $v = \frac{d}{t}$

The helicopter has to travel 1200 miles (distance) in 7 hours (time). Then we can replace the values in the velocity formula.

$v = \frac{1200 }{7 } \\v = 171,42 m/h$

Therefore the helicopter speed is equal to 171 m/h

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A carbon fiber car bumper is hit by another car with the stress of 106,483 Pa. If carbon fiber has a Young's modulus of 228 x 10

hoa [83]

Answer:

Stress = 4.67 * 10^-7 N/m²

Explanation:

Young's modulus of the material = Stress/Strain

Given

Young's modulus = 228 x 10^9 Pa

Stress = 106,483 Pa

Required

Strain

From the formula;

Strain = Stress/Young modulus

Strain = 106,483 /228 x 10^9

Stress = 4.67 * 10^-7 N/m²

3 0

2 years ago

14. The average speed of a car was 60 m/s by the time it reached the finish line. The car moved in a straight line and traveled

Leni [432]

7.5m. Have a good day :)

4 0

2 years ago

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

2 years ago

60,000 is 100 time as much as

saw5 [17]

60,000 is 100 times as much as 600

5 0

2 years ago

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

$R = \frac{2.525}{1000}\times 50$

$R = 0.126 ohm$

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

$V = iR$

$V = 11 \times 0.126$

$V = 1.4 Volts$

so most appropriate answer in given options is

A. 1.8 Volts

8 0

2 years ago