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3 years ago

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A plane is traveling North at 80km/hr into a 20 km/hr headwind (South). What is the plane's resultant velocity?

2 answers:

frosja888 [35]3 years ago

8 0

Answer:

60km/hr

Explanation:

R=80 - 20

=60km/hr, north

Alexxandr [17]3 years ago

5 0

Answer:

Consider our plane flying at a speed of 80 km/hr into a 40 km/hr crosswind. Its speed, relative to the ground, can be calculated by adding the two vectors as shown on the right. What happens to the speed of the plane as the crosswind increases in magnitude? Look at the animation on the right. It is obvious that the speed of the plane, relative to the ground, increases in magnitude.

Explanation:

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If 1.3 grams of air has a volume of 1000cm ^3, what is the density of air?

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Density is equal to mass/volume, so in your case:
density=1.3/1000=0.0013 g/cm^3

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A locomotive with two carriages drives out of the station. The locomotive has a mass of 3.0 tonnes, and each of the two wagons h

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c) Calculate the magnitude of the forces acting on the middle carriage during acceleration.

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Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

frez [133]

Answer:

<em>1.49 x </em> $10^{11}$ <em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

$T^{2}$ ∝ $R^{3}$

<em>to remove the proportionality sign we introduce a constant</em>

$T^{2}$ = k $R^{3}$

k = $\frac{T^{2} }{R^{3} }$

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x $10^{11}$

for mars, constant k will be

k = $\frac{687^{2} }{(2.279*10^{11}) ^{3} }$ = 3.987 x $10^{-29}$

For Earth, orbital period T is 365 days, therefore

$365^{2}$ = 3.987 x $10^{-29}$ x $R^{3}$

$R^{3}$ = 3.34 x $10^{33}$

R =<em> 1.49 x </em> $10^{11}$ <em></em>

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3 years ago

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Answer:

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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

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3 years ago