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Nookie1986 [14]
3 years ago
6

a plane flying due east at 395 km/h, is hit by wind blowing at 55 km/h toward the west. what is the resultant velocity of the pl

ane?
Physics
2 answers:
Aleonysh [2.5K]3 years ago
6 0
In this case, you simply use subtraction to find out the velocity.  If the plane is flying at 395 km/h and is being blown by 55 km.h wind the other way, the velocity of the plane is 395 - 55.

395 km/h - 55 km/h = 340 km/h

So, the final velocity of the plane is 340 km/h east.
FrozenT [24]3 years ago
4 0

I think this is a situation where it's pretty important to mention the reference
frame in which the velocity is being measured.  You've got two different reference
frames going on here.

-- If the plane is flying east at 395 km/hr with respect to the ground, then
that is its velocity with respect to the ground ... 395km/h east.  If in addition
we know that the air is moving 55 km/h west with respect to the ground,
then the plane's velocity with respect to the air around it must be 450 km/h .

-- If the plane is flying east at 395 km/hr with respect to the air around it,
and the air in turn is moving 55 km/h west with respect to the ground, then
the plane's velocity is still 395 km/h through the air (its 'air speed'), and
340 km/h with respect to the ground (its 'ground speed').

The 'ground speed' is the vector sum of the 'air speed' and the wind speed.
There are special plastic devices made that a solo pilot can use to do this
calculation with one hand while he's flying.  He can read his air speed from
a gauge in his plane, and he gets wind speed and direction for his location
by radio from air traffic control centers or flight service stations along his route.
Then he has to calculate what direction he should point the nose of his plane
in order to proceed along the ground in the direction he wants to go.
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3 years ago
A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00
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Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

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The planes’ acceleration from A to B is 500m/s^2

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3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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