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Leya [2.2K]
3 years ago
14

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Physics
1 answer:
stellarik [79]3 years ago
5 0

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft

Also, at any instant t

\Rightarrow l^2=x^2+y^2

differentiate w.r.t.

\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s

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10. Convert the following:<br> a. 37.4 mL into ML<br> b. 689 km/hr into m/s<br> c. 34.5 m² into mm²
Snezhnost [94]

A. When we convert 37.4 mL to ML, the result obtained is 3.74×10¯⁸ ML

B. When we convert 689 km/hr to m/s, the result obtained is 191.39 m/s

C. When we convert 34.5 m² to mm², the result obtained is 3.45×10⁷ mm²

<h3>A. How to convert millimeters (mL) to megaliter (ML)</h3>
  • Volume (mL) = 37.4 mL
  • Volume (ML) =?

1 mL = 1×10¯⁹ ML

Therefore,

37.4 mL = 37.4 × 1×10¯⁹

37.4 mL = 3.74×10¯⁸ ML

Thus, 37.4 mLis equivalent to 3.74×10¯⁸ ML

<h3>B. How to convert 689 km/hr to m/s</h3>

Conversion scale

3.6 Km/hr = 1 m/s

Therefore,

689 km/hr = 689 / 3.6

689 km/hr = 191.39 m/s

Thus, 689 km/hr is equivalent to 191.39 m/s

<h3>C. How to convert 34.5 m² to mm²</h3>

Conversion scale

1 m² = 1×10⁶ mm²

Therefore,

34.5 m² = 34.5 × 1×10⁶

34.5 m² = 3.45×10⁷ mm²

Thus, 34.5 m² is equivalent to 3.45×10⁷ mm²

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

6 0
2 years ago
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
ki77a [65]
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.


7 0
3 years ago
German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ???
fredd [130]

Answer:

\Delta x = 5.47 \times 10^{-9} m

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

\Delta x \times \Delta P = \frac{h}{4\pi}

so here we know that

\Delta v = 0.01 \times 10^6 m/s

m = 9.11 \times 10^{-31} kg

so we have

\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}

\Delta x = 5.47 \times 10^{-9} m

5 0
3 years ago
Pls help ASAP!! Thank you
GaryK [48]

Answer:

Explanation:

The answer you should pick is A. Two magnets will attract each other when their poles are opposite and they are close together.

5 0
3 years ago
Read 2 more answers
What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?
stiks02 [169]

p=mv

p=0.005kg×3.0m/s

p= 0.015kgm/s

4 0
3 years ago
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