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3 years ago

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In the nucleus of an atom, two protons are separated are by a distance of 1*10^-15m. What is the magnitude of the electric force

between them?
A. 115N
B. 720N
C. 142N
D. 230N

1 answer:

xenn [34]3 years ago

4 0

Answer:

D. 230 N

Explanation:

The magnitude of the electric force between the two protons is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 = q_2 = 1.6\cdot 10^{-19} C$ is the charge of each proton

$r=1\cdot 10^{-15} m$ is the distance between the two protons

Substituting numbers into the formula, we find

$F=(9\cdot 10^9 N m^2 C^{-2}) \frac{(1.6\cdot 10^{-19} C)^2}{(1\cdot 10^{-15} m^2)^2}=230.4 N \sim 230 N$

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In a chemical reaction, molecules of hydrogen gas (H2) react with molecules of oxygen gas (O2) in a sealed reaction chamber to p

11Alexandr11 [23.1K]

Answer:

Option (B) is correct.

Explanation:

Given that the molecules of hydrogen gas ( $H_2$ ) react with molecules of oxygen gas ( $O_2$ ) in a sealed reaction chamber to produce water ( $H_2O$ ).

The governing equation for the reaction is

$2H_2 +O_2 \rightarrow 2H_2O$

From the given, the only fact that can be observed that 2 moles of $H_2$ and 1 mole of $O_2$ reacts to produce 2 moles of $H_2O$ .

As the mass of 1 mole of $H_2 = 2$ grams ... (i)

The mass of 1 mole of $O_2 = 32$ grams ...(ii)

The mass of 1 mole of $H_2O = 18$ grams (iii)

Now, the mass of the reactant = Mass of 2 moles of $H_2$ + mass 1 mole of $O_2$

$= 2 \times 2 + 32$ [ using equations (i) and (ii)]

$=4+32 = 36$ grams.

Mass of the product = Mass of 2 moles of $H_2O$

$=2\times 18$ [ using equations (iii)]

=36 grams

As the mass of reactants = mass of the product.

So, mass is conserved.

Hence, option (B) is correct.

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3 years ago

A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an

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Answer:

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Explanation:

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Why do we use atomic models?

vlabodo [156]

Models help us to understand systems and their properties

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2 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

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To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

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$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

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3 years ago

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

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3 years ago