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stellarik [79]
3 years ago
14

How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.0

0×10−15 m (a typical nuclear distance)?
Physics
1 answer:
Deffense [45]3 years ago
6 0

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

W=q \Delta V

where q=1.6 \cdot 10^{-19}C is the charge of the proton and \Delta V is the potential difference between the final position and the initial position of the proton. To calculate this \Delta V, we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

V=k\frac{Q}{r}

where k=9.0 \cdot 10^9 N m^2 C^{-2} is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

V_i = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{2.0 \cdot 10^{-10}}=7.2 V

V_f = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{3.0 \cdot 10^{-15}}=4.8 \cdot 10^5 V

Therefore, the work done is

W=q \Delta V=(1.6 \cdot 10^{-19}C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^{-14} J

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What is the difference in average speed between two cars, one that traveled 160 kilometers in 5
Monica [59]

Answer:

4 km/h

Explanation:

160/5 = 32

140/5 = 28

32-28= 4

4 0
4 years ago
1) A box is 16m long, 28m broad and 10m high.what is the volume of 100 similar boxes?2)When was S.I unit introduced? (3)0How man
Romashka-Z-Leto [24]

Explanation:

1)a=16m

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c=10m

volume of one box: V=16*28*10=4480m³

volume of 100 boxes: 4480*100=448000m³

2) 1960

3) 7: meter-length, kilogramme-mass, second-time, ampere-electric current, kelvin-temperature, candela-luminous intensity, mole-amount of substance

4)Area: S=a*b

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3 0
4 years ago
Suppose a runner completes a 10k (10.0 km) road race in 41 minutes and 25 seconds. what is the runner's average speed in miles p
velikii [3]

Answer;

9 miles per hour.

Solution;

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41 min = 2460 seconds (+ 25 seconds) = 2485/ 60 = 41.417 minutes (0.6903 hours).

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But 1 mile = 1.60934 km

Thus; 14.4869 kilometers = 9.0017 miles.

So the runner's average speed is roughly 9 miles per hour.

8 0
4 years ago
A car is traveling 11.4 m/s. A truck having four times the mass of the car is traveling at 22.7 m/s. How many times greater is t
Marta_Voda [28]
Sice the mass are four times the car the relationship between both is inverse so the kinetic is 4 times less
6 0
3 years ago
How could you increase the amount of kenitic energy
Fiesta28 [93]

Answer:

As You know kinetic Energy is equal (mv^2)/2

You can increase it by increasing the mass(keeping the velocity)

Or You can add velocity

Explanation:

8 0
3 years ago
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