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Komok [63]
3 years ago
11

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

Physics
1 answer:
Katen [24]3 years ago
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

\frac{1}{2}mv^2 = \frac{1}{2}kX^2

Where,

m=mass

v=velocity

k=spring constant

x=amount of compression

Re-arrange the equation to find the velocity we have,

v^2 = \frac{kX^2}{m}

v^2 = \frac{10(0.25)^2}{50}

v = \sqrt{0.0125}

v = 0.11m/s

Therefore the maximum speed of the bananas is 0.11m/s

8 0
3 years ago
A puck of mass m 0.085 kg is going in a circle on a horizontal frictionless surface. It is held in its orbit by massless string
attashe74 [19]

Answer:

Explanation:

Magnitude of tension

= centripetal force

= m ω²R

= m (2π/T)² R , T is time period of revolution.

= .085 x (2 x 3.14 / .55)² x .84

= 9.3  N .  

8 0
3 years ago
5. A store clerk moved a 4.4-kg box of soap without acceleration along a shelf by pushing it with a horizontal force of magnitud
lesantik [10]

The box moved 0.73 m

<u>Explanation:</u>

Given data,

Magnitude 8.1 N Work on the box 5.9 J Clerk move a 4.4 Kg

Distance= work done on the box / Horizontal force on the magnitude

              = 5.9/8.1

               =0.728

Distance =0.73 m

The box moved 0.73 m

5 0
3 years ago
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