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3 years ago

If an aircraft is flying at 400 miles/hour and encounters a head wind of 40 miles/hour, what is the resultant ground speed?

1 answer:

8 0

Look at it this way:

Every time the plane flies through 400 miles of air,
the air moves it 40 miles backwards.
So it only covers 360 miles of ground in an hour.

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A ball was rolling downhill at 2 m/s. After 5s, it was rolling at 90 m/s. What is its acceleration?

olga55 [171]

Answer:

17.6 m/s²

Explanation:

Given:

$v_{f}$ = 90 m/s (final velocity)

$v_{i}$ = 2 m/s (initial velocity)

Δt = 5s (change in time)

The formula for acceleration is:

$a_{avg}$ = Δv / Δt

We can find Δv by doing

Δv = $v_{f}$ - $v_{i}$

Replace the values

Δv = 90m/s - 2m/s

Δv= 88m/s

Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.

$a_{avg}$ = Δv / Δt

$a_{avg}$ = $\frac{88m/s}{5/s}$

acceleration = 17.6 $m/s^{2}$

4 0

3 years ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

Natasha_Volkova [10]

Answer

given,

distance of first satellite = 48,000 Km

distance of second satellite = 64,000 Km

orbital period = 6.39 day

Using equation of time period

$T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_{pluto}}}$

now, from the above equation we can say that only variable is Time period and r is the radii of orbit.

from the first satellite

$\dfrac{T_{charon}}{r^{3/2}_{charon}}=\dfrac{T_{sat1}}{r^{3/2}_{sat1}}$

$T_{sat1}=\dfrac{T_{charon}\ r^{3/2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (48000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=24.5\ days$

for second satellite

$T_{sat2}=\dfrac{T_{charon}\ r^{3/2}_{sat2}}{r^{3/2}_{charon}}$

$T_{sat1}=\dfrac{6.39\times (64000)^{3/2}}{19600^{3/2}}$

$T_{sat1}=37.7\ days$

7 0

3 years ago

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 25 cm on a spring with a spring con

KiRa [710]

Answer:

0.11m/s

Explanation:

To solve the exercise it is necessary to apply the concepts related to the conservation of both: kinetic and spring energy(Elastic potential energy), in this way

Kinetic Energy = Elastic potential energy

$KE = SE$