That's going to depend on the mass of the crate. They'll be able to push a crate full of pingpong balls really high, but they may not be able to even budge the crate at all if it's full of bricks.
F= ma
F= (600/-10) -10
F= 580n
At least I think that’s the answer
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.
Answer:
BW = 100 rad/s
wlow = 452.49 rad/s
whigh = 552.49 rad/s
V(jwlow) =1414.21 < 45°V
V(jwhigh) =1414.21 <-45°V
Explanation:
To calculate bandwidth we have formula
BW = 1/RC
BW = 1/ 1000x10x10^¯6
BW = 100 rad/s
We will first calculate resonant frequency and quality factor for half power frequencies.
For resonant frequency
wo = 1/(SQRT LC)
wo = 1/SQRT 400×10¯³ × 10×10^¯6
wo = 500 rad/s
For Quality
Q = wo / BW
Q = 500/100
Q = 5
wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]
wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]
wlow = 452.49 rad/s
whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]
whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]
whigh = 552.49 rad/s
We will start with admittance at lower half power frequency
Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)
Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)
Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³
Y(jwlow) = (1-j).10¯³ S
Voltage across the network is calculated by ohm's law
V(jwlow) = I/Y(jwlow)
V(jwlow) = 2/(1-j).10¯³
V(jwlow) = 1414.2 < 45°V
Now we will calculate the admittance at higher half power frequency
Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)
Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)
Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³
Y(jwhigh) = (1+j).10¯³ S
Voltage across network will be calculated by ohm's law
V(jwhigh) = I/Y(jwhigh)
V(jwhigh) = 2/(1+j).10¯³
V(jwhigh) = 1414.2 < - 45°V
