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2 years ago

What does it mean to be skeptical of health fraud

Physics

2 answers:

vesna_86 [32]2 years ago

7 0

Something that is falsely advertised to help different health issues. An example could be Eyeball massagers

snow_tiger [21]2 years ago

6 0

Answer: yea ma’am I’m sorry but you still

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What is the split atom called?

son4ous [18]

If you're talking about the <em>splitting</em> of an atom, the process is called Nuclear Fission.

3 0

3 years ago

Read 2 more answers

A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra

Sonja [21]

Answer:

Intensity, $I=1.101\ W/m^2$

Explanation:

Power of the light bulb, P = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

$I=\dfrac{40\ W}{4\pi (1.7\ m)^2}$

$I=1.101\ W/m^2$

So, the intensity of light is $1.101\ W/m^2$ .

6 0

3 years ago

Maggie is using her cat in an experiment on animal intelligence. Which of the following statements is true?

Kamila [148]

1. If Maggie gives her cat an unfair advantage, her experimental results will be biased.

Explanation:

Maggie using her cat in the experiment to test for intelligence gives the cat an unfair advantage, her experimental results will be biased.

Due to her emotional attachment with the cat, the experimental results will be skewed to portraying her cat as intelligence.
This is not a good experiment to carry out.
Such an experiment should be carried out with an unknown cat.

learn more:

Experiment brainly.com/question/1621519

#learnwithBrainly

6 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

What is the Mirror equation? What does it look like?

kipiarov [429]

Answer:

1/f = 1/D' + 1/D

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (h^i) and object height (h^o). The magnification equation is stated as follows:

M= H^i/H^o = D^i/D^o

7 0

2 years ago