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Maslowich
2 years ago
6

(For example, if you used a lunch meat that claimed to be 97% fat-free, were just 3% of its calories provided by fat?) If the ma

nufacturer made such claims, how do you explain the difference between the manufacturer's claims and your findings?
Salmon
Physics
1 answer:
klasskru [66]2 years ago
3 0

The difference between the manufacturer's claims and your findings is, the manufacturer is concerned about amount of fat in the meat while you are concerned about calories.

<h3>Difference between your findings and the manufacturers claim</h3>

Your findings concluded that only fat can provide calories needed in the body, perhaps other classes of food or nutrients in the meat can provide calories as well.

In the manufacturer's claim there are about 3% fat in the lunch meat and doesn't necessary mean that all the calories needed is the 3% fat.

Thus, there are lots of difference between your findings and the manufacturer's claim. The manufacturer is concerned about amount of fat in the lunch meat while you are concerned about calories in the meat.

Learn more about claims here: brainly.com/question/2748145

#SPJ1

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In one town, there are several natural gas drilling sites. Scientists warn town officials that the health of the people could be
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Answer:

C. community

Explanation:

Community consists of a local residence, like a town.

7 0
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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
dedylja [7]
<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.

\theta=30\° the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at 30\°:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))   (2)

\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m   (3)

Now, the initial energy E_{o}=400keV=400(10)^{3}eV of the photon is given by:

 E_{o}=\frac{h.c}{\lambda_{o}}    (4)

From this equation (4) we can find the value of \lambda_{o}:

\lambda_{o}=\frac{h.c}{E_{o}}    (5)

\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}    

\lambda_{o}=3.102(10)^{-12}m    (6)

Knowing the value of \Delta \lambda and \lambda_{o}, let's find \lambda':

\Delta \lambda=\lambda' - \lambda_{o}

Then:

\lambda'=\Delta \lambda+\lambda_{o}  (7)

\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m  

\lambda'=3.427(10)^{-12}m  (8)

Knowing the wavelength of the scattered photon \lambda'  , we can find its energy E' :

E'=\frac{h.c}{\lambda'}    (9)

E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}    

E'=362.063keV    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron E_{e}, we have to calculate all the energy lost by the photon, which is:

E_{e}=E_{o}-E'  (11)

E_{e}=400keV-362.063keV  

Finally we obtain the energy of the recoiling electron:

E_{e}=37.937keV  

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3 years ago
An electric iron connected to a 110 V source draws 9 A of current. How much heat (in joules) does it generate in a minute?
meriva
Q = 110V x 9A x 60s = J
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A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s
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Answer:

0.4rad/s²

Explanation:

Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m

a = ∆w/t

a = (w2 - w1)/t

a = (2 - 4)/5 = -2/5 =

a = - 0.4rad/s²

The -ve sign indicates a deceleration in the motion

Good luck

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State ohm’s law Q.34.7
defon

Ohm's law states that: V=RI

Explanation:

Ohm's law states that in a conductor, the potential difference across the conductor is directly proportional to the current flowing through it. Mathematically,

V \propto I

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V is the potential difference

I is the current

The constant of proportionality is called resistance (R), and it gives a measure of "how much the conductor opposes" to the flow of current. Therefore Ohm's law can be rewritten as

V=RI

where R is the resistance. By rewriting the equation as

I=\frac{V}{R}

we see that the larger the resistance, the lower the current in the conductor.

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