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3 years ago

If an airplane travels a distance of 540 km in 2.5 hours, what is its average speed in m/s?

Physics

1 answer:

VashaNatasha [74]3 years ago

6 0

Answer:

Great question! To solve this problem you just need a bit of simple math. The short answer is 60 m/s

Explanation:

To solve this, you must first set up your equation.

540km/2.5hr * 1000m/3600s

You would set it up this way because there are 1000 meters in a km and 3600 seconds in an hour. This

540000m/9000s

This simplifies down to

60m/s

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What is the force on a 250 kg elevator that is falling freely at 9.8 m/sec2?

Viefleur [7K]

Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

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3 years ago

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B. my town is powered by electricity that is generated by the energy from the flow of water through a large dam.

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Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

F = Magnetic force = 4.11 N

$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

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3 years ago

A 5000 g toy car starts from rest and moves a distance of 300 cm in 3 s under the action of a single constant force. Determine t

sveticcg [70]

Answer:

3.33 N

Explanation:

First, find the acceleration.

Given:

Δx = 3 m

v₀ = 0 m/s

t = 3 s

Find: a

Δx = v₀ t + ½ at²

3 m = (0 m/s) (3 s) + ½ a (3 s)²

a = ⅔ m/s²

Use Newton's second law to find the force.

F = ma

F = (5 kg) (⅔ m/s²)

F ≈ 3.33 N

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3 years ago

approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos

Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as: $v_{av} = \frac{x - x_{0} }{t - t_{0} }$

where t = final time

t₀ = initial time

v = final speed

v₀ = initial time

where x denotes the position, then

$v_{ave} = \frac{v+v_{0} }{2}$

where v = $\frac{dx}{dt}$ and dx = change in distance with respect to time.

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4 years ago