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Arturiano [62]
3 years ago
5

How much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 km/hr for a dista

nce of 4,700km?
Also, if the airplane then turns around and returns to its original gate, what is its velocity?
Physics
1 answer:
andrezito [222]3 years ago
3 0

Time = (distance) / (speed)  =  (4,700 km) / (790 km/hr)  =  5.95 hours


Velocity =

           (distance from the start-point to the end-point)
divided by
           (time for the trip) .

If the airplane went 4,700 km away, then turned around and returned
to its starting point, then the distance from the start-point to the end-point
is zero, and the average velocity for the trip is zero.

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A horse travels 80 meters south in 20 seconds. What is its average velocity?
AURORKA [14]

Average velocity is a vector unit (i.e. includes magnitude <em>and </em>direction) calculated by working out distance ÷ time:

80 metres ÷ 20 seconds = 4 metres/seconds (m/s)

Therefore, your final answer is C. 4 m/s south.

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A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
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Explanation:

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As initial speed is zero, therefore;

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Answer:

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Explanation:

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