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3 years ago

Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?

2 answers:

4 0

Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity.

shutvik [7]3 years ago

4 0

Depends on the object. Is it round? Is it a cube? Air resistance all depends on what kid of surface the object shows when it is falling, thus you have parachutes to stop your velocity from making you go splat.

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A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

What is the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind?

salantis [7]

We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is

X=0m/s

From the question we are told that

mosquito flying 2 m/s

against a 2 m/s headwind

Generally

The speed over the ground is the Flight Speed minus resistance speed

Generally the equation for the speed over the ground is mathematically given as

X=Flight Speed-resistance speed

Therefore

X=2-2

X=0m/s

For more information on this visit

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8 0

2 years ago

Read 2 more answers

How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?

saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

5 0

2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers

Is it possible for an object that has a constant negative acceleration to change the direction in which it is moving? Explain wh

lozanna [386]

Yes! I think there are two ways you could go with this answer: 1) Acceleration is the change in velocity over time, it can be negative or positive. If you have an object that is already moving forwards in a straight line and give it a constant negative acceleration, it will slow down and then start going in reverse. 2)Velocity is a vector, meaning it has both magnitude and direction. In the example above, the acceleration is due to a change in magnitude, or speed (from +ve to -ve) but not a change in direction. Something that has constant speed but is changing direction is also accelerating (like something that is orbiting). You could use the earth as an example, which is constantly accelerating due to moving in a circle around the sun. At any time in the year you can say that in half a year's time the earth's direction will be reversed.

4 0

2 years ago