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nexus9112 [7]
3 years ago
12

A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc

e! 
Physics
1 answer:
ki77a [65]3 years ago
7 0
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
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:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

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.

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

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

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

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
Why are viruses hard to fight
ki77a [65]

Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.

Explanation:

7 0
2 years ago
A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,
Natasha_Volkova [10]

Answer:

the average force 11226 N  

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest  (Vo=0)

             

           Vf² = Vo² - 2gY

            Vf² = 0 - 2 9.8 7.69 = 150.7

            Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam.  They also give us the distance it travels before stopping, let's calculate the time

         

            Vf = Vo - g t

             0 = Vo - g t

             t = Vo / g

             t = 12.3 / 9.8

             t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

                 I = Δp

                F t = pf-po

The amount of final movement is zero because the system stops

                F = - po / t

                F = - mv / t

                F = - 1150 12.3 / 1.26

                F = -11226 N

This is the average force exerted by the stack on the vean

7 0
2 years ago
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marysya [2.9K]
Answer:

The answer is D. density.
8 0
2 years ago
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