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3 years ago

Classify each possible hypothesis about a medicinal aloe vera plant as falsifiable or non-falsifiable.

Physics

1 answer:

pychu [463]3 years ago

7 0

Answer:

Aloe vera is a succulent plant species ( with fleshy water storing tissue) belonging to the genus Aloe.

It is used in many products ( including beverages, cosmetic products as gel, medicinal ointments). Its juice is quite bitter in taste. It is known to be the best skin moisturizer and can be used for healing wounds by enhancing the renewal of damaged cells. It is also used as an anticarcinogen, thus reduces the chances of cancer.

Thus, the given hypothesis can be classified as falsifiable ( which is proven wrong) or non-falsifiable ( which is proven right) in the following way-

Falsifiable-

Aloe vera juice tastes better than carrot juice.

Non Falsifiable-

Aloe vera gel is the best natural skin moisturizer.

Aloe vera gel can heal wounds by boosting cell renewal.

Drinking aloe juice can reduce the risk of lung cancer.

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A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

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Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

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Distance, s = 12 m

Time, t = 1.2 s

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m a = q E.......(1)

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a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

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$q=\dfrac{ma}{E}$

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3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

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So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

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