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Anit [1.1K]
2 years ago
8

Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th

eir cross-sectional areas were different. If axial force is applied to each rod such that they all change length by the same amount, which rod experienced the largest force
Physics
1 answer:
Shkiper50 [21]2 years ago
4 0

The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

following formula;

\Delta L = \dfrac{F \cdot L}{A \cdot E}

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

a given force reduces as the cross sectional area of the rod increases.

Therefore, the force required to extend the length of a rod by a specific

amount increases as the cross sectional area of the rod increases,

indicating that the rod with the largest cross sectional area require the

most force and therefore, experiences the largest force.

The rod that experiences the largest force is the rod with the largest cross

sectional area, which is <u>rod B</u>

Learn more here:

brainly.com/question/12937199

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Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0
3 years ago
Delilah does 170 Joules of work in 30 seconds. Adam does 260 Joules of work in 20 seconds. Who was more powerful?​
sp2606 [1]

Power = \frac{Work}{Time}

Delilah: 170J/30s = 5.66 W

Adam: 260J/20s = 13 W

6 0
3 years ago
A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t
Arisa [49]

A) 2.4\cdot 10^{-16}kg

The radius of the oil droplet is half of its diameter:

r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m

Assuming the droplet is spherical, its volume is given by

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3

The density of the droplet is

\rho=885 kg/m^3

Therefore, the mass of the droplet is equal to the product between volume and density:

m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg

B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

3 0
3 years ago
Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti
Gwar [14]
After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it.  So only  (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter,  96.6%  of what entered it emerges from it, and
that's  96.6%  of  96.6%  of the original signal that entered the beginning
of the fiber.

==>  After 2 meters, the intensity has dwindled to  (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==>  After  'x'  meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
                                                     <em>(1,500) · (0.966)^x</em>
lumens of light remaining.
 
=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say  "15dB per 100 meters".

What does that mean ?    Break it down:  15dB in 100 meters is <u>0.15dB per meter</u>.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is  3.4% .  We applied
super high mathematics to that and calculated that  96.6% remains, or  0.966.

Look at this  ==>      10 log(0.966) =  <em><u>-0.15</u>  </em>  <==  loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in  'x'  meters of
fiber cable, doesn't have to mess with raising numbers to powers.  All he has to
do is say ...

--  0.15 dB loss per meter

--  'x' meters of cable

--  0.15x dB of loss.

If  'x' happens to be, say,  72 meters, then the loss is  (72) (0.15) = 10.8 dB .

and  10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083  =  <em>8.3%</em>  <== <u>That's</u> how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.
5 0
3 years ago
A vibrating tuning fork is held over a water column with one end closed and the other open. As the water level is allowed to fal
DiKsa [7]

Answer:

630 Hz.

Explanation:

As we are considering the one end open pipe. So for the sound wave there will be a pressure node at the open end of the tube as at that place the molecules can not move back and forth. However on the closed end there will be a flow node as the water molecules their are moving back and forth. So it will produces the resonance at the positions 1/4, 3/4.......

we can find the wavelength by multiplying the levels distance by 2.

λ = 2 × 0.27 m = 0.54

f = Vs/λ

  = 340/0.54

  = 630 Hz

6 0
3 years ago
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