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oksian1 [2.3K]
3 years ago
14

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
Physics
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.

8 0
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
svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

Download pdf
3 0
3 years ago
Why do we use atomic models?
vlabodo [156]
Models help us to understand systems and their properties
3 0
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
Mademuasel [1]

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

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

The given data are given with respect to known constants, for example the mass of the sun is

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)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
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.

4 0
3 years ago
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