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arlik [135]
4 years ago
15

An airplane moves through the air at a constant speed. The engines’ thrust applies a force in the direction of motion, and this

force is equal in magnitude and opposite in direction to the drag force. Reducing thrust will cause the plane to fly at a slower—but still constant—speed. Explain why this is so.
Physics
1 answer:
lord [1]4 years ago
8 0

Explanation:

This is because the drag force suffered by the aircraft is proportional to the speed at which it travels. The thrust of the engines prints a speed to the plane and this speed prints a drag force, always reaching an equilibrium point of these two forces where the speed of the plane is constant and the acceleration is equal to zero.

Therefore, by reducing the thrust, the drag force is greater and the plane begins to decrease its speed, until it reaches the point where the new drag force is matched with the new thrust force, giving it a new final speed , without acceleration.

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A roller coaster is stopped on a track. When the engineer presses a launch button on the coaster, the coaster moves forward. Explain this change in terms of balanced and unbalanced forces.
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Fill in the blanks.<br>In what way are speed and velocity different?​
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A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true
Fantom [35]

Answer:

(d) not enough info

Explanation:

because it doesn't specify where the strings are attached

if it was the two ends of the rod then T1 would be equal to T2

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3 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

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p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
A motorcycle, which has an initial linear speed of 9.7 m/s, decelerates to a speed of 4.0 m/s in 4.4 s. Each wheel has a radius
Morgarella [4.7K]

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

6 0
3 years ago
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