If an airplane is traveling at 500 mph and a fly is flying forward inside the airplane is the fly traveling faster than the airplane?

Answers

Answer 1

Answer: Relative to the Earth, yes; The inside of the airplane acts as its own kind of "world", in which the standard would be moving 500 mph relative to the Earth. As the fly is moving forward inside, you would add the 500 mph of the plane plus the speed of the fly relative to the plane to get the speed of the fly relative to the Earth.

Answer 2

Answer:
Any motion is only measurable by comparing it with something else.
In the same frame of reference in which the plane is traveling at 500 mph,
yes, the fly is moving faster than that. But from the point of view of a passenger
on the plane, no it isn't.

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A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force of static friction f on the block?A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Answers

When the block is at rest, the static frictional force is equal to the horizontal component of the block's weight (F = mgsin(θ)).

The static frictionalforce on the body at rests is determined by applying Newton's second law of motion.

F = ma

where;

F is applied force on a body
m is the mass of the body
a is the acceleration of the body

If the block is at rest, then the net horizontal force on the block is zero.

$\Sigma F_x = 0\n\nmg sin(\theta ) - F_s = 0\n\nF_s = mg sin(\theta)$

Thus, when the block is at rest, the static frictional force is equal to the horizontal component of the block's weight.

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Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that

[tex]f=mgsin \theta[tex]

Therefore, option B is the correct option.

I need help ASAP.The average time that it takes for the car to travel the first 0.25m is_____

The average time to travel just between 0.25m and 0.50m is_____

Given the time taken to travel the second 0.25m section, the velocity would be____m/s

Answers

1. Average time for the first 0.25 m: 2.23 s

Explanation:

The average time that it takes for the car to travel the first 0.25 m is given by the average of the first three measures, so:

$t=(2.24s+2.21s+2.23s)/(3)=2.227 s=2.23 s$

2. Average time to travel between 0.25 m and 0.50 m: 0.90 s

Explanation:

First of all, we need to calculate the time the car takes to travel between 0.25 m and 0.50 m for each trial:

t1 = 3.16 s - 2.24 s = 0.92 s

t2 = 3.08 s - 2.21 s = 0.87 s

t3 = 3.15 s - 2.23 s = 0.92 s

So, the average time is

$t=(0.92 s + 0.87 s + 0.92 s)/(3)=0.903 s=0.90 s$

3. Velocity in the second 0.25 m section: 0.28 m/s

Explanation:

The average velocity in the second 0.25 m section is equal to the ratio between the distance covered (0.25 m) and the average time taken (0.90 s):

$v=(d)/(t)=(0.25 m)/(0.90 s)=0.28 m/s$

Answer:

1. 2.23

2. 0.90

3. 0.28

In a series circuit, the current drops across each resistance.
a. True
b. False

Answers

That's false. In a series circuit, the current is the same
at every point all around the circuit.

It's the voltage that drops across each resistance.

The odometer gauge indicates the ________________.

Answers

The odometer gauge indicates the ________________.

the distance traveled.

Carlos gets tired of pushing and instead begins to pull with force Fpull at an angle to the horizontal.The block slides along the rough horizontal surface at a constant speed. A free-body diagram for the
situation is shown below. Blake makes the following claim about the free-body diagram:
Blake: “The velocity of the block is constant, so the net force exerted on the block must be zero.
Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied
force Fpull.”
What, if anything, is wrong with this statement? If something is
wrong, identify it and explain how to correct it. If this statement is
correct, explain why.

Answers

Answer:

The wrong items are;

1) The normal for FN equals the weight Fmg

2) The force of friction, Ff, equals the applied force Fpull

The corrected statements are;

1) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ)

Explanation:

The given statement was;

The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull

By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;

The normal force is equal to the weight less the vertical component of the applied force;

That is we have, FN = Fmg - Fpull × sin(θ)

The friction force similarly, is equal to the horizontal component of the applied force;

Ff = Fpull × cos(θ)

The wrong items are therefore as follows;

1) The normal for FN equals the weight Fmg

1 i) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction, Ff, equals the applied force Fpull

2 i) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ).

Final answer:

While Blake's statement about the normal force is correct, his claim about the applied force and friction force is partially accurate. In reality,the horizontal component of the applied force should equate to the friction force for the block to maintain a constant velocity.

Explanation:

Blake's claim that the normal force FN equals the weight Fmg is correct as these forces balance each other in the vertical direction. However, his claim that the force of friction Ff equals the applied force Fpull is only partially accurate. In reality, the horizontal component of Fpull (i.e., Fpull * cos(θ)) should equate to the friction force Ff, to maintain the constant velocity (the block is not accelerating). The vertical component of Fpull (i.e., Fpull * sin(θ)) reduces the effective weight of the block and thereby, the normal force.

To correct Blake's claim, the normal force FN is equal to the weight of the block minus the vertical component of the applied force, and the applied force's horizontal component equals the friction force. Hence, this is the correct solution considering both vertical and horizontal components of forces.

Learn more about Force Components here:

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As you (increase, decrease) in altitude, air pressure decreases.

Answers

Answer:

as you increase in altitude, air pressure decreases

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