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Law Incorporation [45]

3 years ago

8

Find the percentage of the total work lost to friction if 28.7 J of work is put into pushing a block up a ramp resulting in 14.2

J of stored potential energy at the top.

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

14.5 joules

50.5 %

Explanation:

Energy initial +work= energy final

28.7 + work = 14.2

friction did - 14.5 joules of work so percantage of work lost is just 14.5/28.7 *100 or 50.5 %

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3 years ago

Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, t

kkurt [141]

Answer:

$t=0.038s$

Explanation:

Project mass m=3.8 kg

Initial speed vi= 0m/s

Final speed vf= 9.3×10³ m/s

Force F=9.3×10⁵N

To find

Time t

Solution

From Newtons second law we know that

∑F=ma

Where m is mass

a is acceleration

We can write this equation as

∑F=m(Δv/Δt)

$=m\frac{v_{f}-v_{i}}{t}$

Rearrange this equation to find time t

So

$t=m\frac{v_{f}-v_{i}}{F}$

Substitute the given values

$t=3.8kg\frac{9.3*10^3m/s-0}{9.3*10^5N} \\t=0.038s$

5 0

3 years ago

A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The

d1i1m1o1n [39]

Answer:

The value to be reported is 5.48V

Explanation:

The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.

The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.

It is expressed as $V_{rms}$ = $\frac{V_{m} }{\sqrt{2} }$

where Vm is the maximum or peak value of the voltage

In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)

$V_{rms}$ = $\frac{7.75}{\sqrt{2} }$

= $\frac{7.75}{1.414}$

= 5.48 V

6 0

3 years ago

An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.

adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

$\Delta x\times \Delta p=\frac{h}{4\pi}$ ...........(1)

where,

$\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

$p=m\times v$

or,

$\Delta p=m\times \Delta v$ .......(2)

Equating 1 and 2, we get:

$\Delta x\times m\times \Delta v=\frac{h}{4\pi}$

$\Delta v=\frac{h}{4\pi \Delta x\times m}$

Given:

m = mass of electron = $9.11\times 10^{-31}kg$

h = Planck's constant = $6.626\times 10^{-34}Js$

radius of atom = $67.0pm=67.0\times 10^{-12}m$ $(1pm=10^{-12}m)$

$\Delta x$ = diameter of atom = $2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m$

Now put all the given values in the above formula, we get:

$\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}$

$\Delta v=4.32\times 10^{5}m/s$

The minimum uncertainty in an electron's velocity is, $4.32\times 10^{5}m/s$

Now we have to calculate the percentage of the average speed.

Percentage of average speed = $\frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100$

Uncertainty in speed = $4.32\times 10^{5}m/s$

Average speed = $1.3\times 10^{6}m/s$

Percentage of average speed = $\frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100$

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

3 0

3 years ago