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Anni [7]
2 years ago
11

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je

t is moving with a speed of 400 m/s.
What is the jet's acceleration, assuming it to be a constant acceleration?


I got :

a =
8.8
m/s2

Is your answer reasonable? Explain.
Physics
1 answer:
jenyasd209 [6]2 years ago
7 0
<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s  

Acceleration, a = ?

Final velocity, v = 400 m/s  

Displacement,s = 4 km = 4000 m

Substituting  

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

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0.36m/s^2

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7 0
2 years ago
If a force does a positive amount of work on an object, does the object's speed
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The speed of the object increases

Explanation:

We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

W is the work done on the object

K_f, K_i are the final and initial kinetic energy of the object, respectively

m is the mass of the object

v is its final speed

u is its initial speed

In this case, the force does a positive amount of work on the object, so

W>0

This also implies that

K_f > K_i

And so

\frac{1}{2}mv^2 > \frac{1}{2}mu^2

And therefore

v>u

which means that the speed of the object increases.

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8 0
3 years ago
Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will
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Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

P\cdot V = n \cdot R_{u}\cdot T (1)

Where:

P - Pressure.

V - Volume.

n - Molar quantity, in moles.

R_{u} - Ideal gas constant.

T - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

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If we know that \frac{V_{2}}{V_{1}} = \frac{1}{3}, then the resulting pressure of the system is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 3\cdot P_{1}

The resulting pressure is 3 times the initial pressure.

4 0
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