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2 years ago

Why can we never prove that a hypothesis is true?

1 answer:

7 0

A theorem can be proven (from axioms or prior theorems), using logic.

A hypothesis can be supported by evidence. The more evidence in support of the hypothesis, the more likely the hypothesis is to be correct. However, you’re always at the mercy of contrary evidence appearing in the future, to reduce the likelihood or even invalidate a hypothesis.

A (mathematical) proof suffers no such vulnerability to future evidence, as long as you hold the axioms of the theory to be true, and as long as there was no flaw in the construction of the proof.

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Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h

lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is 27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus $7=\frac{x}{v}$ ----1

for return trip

$4=\frac{x}{v+27}$ -----2

divide 1 & 2

$\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}$

$7v=4v+4\cdot 27$

$3v=4\cdot 27$

$v=36 mph$

thus $x=7\times 36=252\ miles$

7 0

2 years ago

What does deposition mean in a sedimentary formation

Zinaida [17]

Answer:

Sedimentary rocks are types of rock that are formed by the deposition and subsequent cementation of that material at the Earth's surface and within bodies of water. Deposition means that all the sediments, soil, and rocks are all compressed (tightly pressed into each other) and create sedimentary rocks.

I hope this helps! :)

4 0

2 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

2 years ago

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240

Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

$V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC$

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0

2 years ago

Six artificial satellites circle a space station at a constant speed. The mass m of each satellite, distance L from the space st

damaskus [11]

Answer:

The answer to the question is;

Based on their acceleration the rank of the satellites from largest to smallest is.

B >→ A >→ E >→ C >→ F >→ D.

Explanation:

Acceleration is given by $\frac{v^2}{r}$

Therefore the acceleration for each of the satellite is given by

Satellite A) $\frac{160^2}{5000}$ = 5.12 m/s²

Satellite B) $\frac{160^2}{2500}$ = 10.24 m/s²

Satellite C) $\frac{80^2}{2500}$ = 2.56 m/s²

Satellite D) $\frac{40^2}{10000}$ = 0.16 m/s²

Satellite E) $\frac{120^2}{5000}$ = 2.88 m/s²

Satellite F) $\frac{80^2}{10000}$ = 0.64 m/s²

Therefore in order of decreasing acceleration, from largest to smallest we have

Satellite B) > Satellite A) >Satellite E) >Satellite C)>Satellite F)>Satellite D).

3 0

2 years ago

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