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arlik [135]
3 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]3 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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Why does the frequency of a wave increase as the wavelength decreases
olchik [2.2K]

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

3 0
4 years ago
Read 2 more answers
It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.
Alenkasestr [34]
The work to stretch a spring from its rest position is

               (1/2) (spring constant) (distance of the stretch)²

           E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

                 1700 joules  =  1/2 k (3m)²

1 joule = 1 newton-meter

                 1700 N-m =  1/2 k  (3m)²

Multiply each side by 2:   3400 N-m  =  k · 9m²   

Divide each side by  9m²     k  =  3400 N-m / 9m²

                                                   =  (377 and 7/9)   newton per meter    

7 0
4 years ago
Read 2 more answers
A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0
3 years ago
What does the object below model?
harina [27]

Answer:  c

Explanation: it is c because i used my brain to answer it

3 0
3 years ago
what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a d
elena55 [62]

Answer:

from \: newton \: law \: of \: gravitation \\ F =  \frac{GMm}{ {r}^{2} }  \\ G = 6.67 \times  {10}^{ - 11}  \\ M = 0.5 \: kg \\ m = 0.33 \: kg \\ r = 0.002 \: m \\ substitute \\ F =  \frac{(6.67 \times  {10}^{ - 11}) \times (0.5) \times (0.33)) }{ {(0.002)}^{2} }  \\  = 2.75 \times  {10}^{ - 6} N

4 0
3 years ago
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