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Snezhnost [94]
3 years ago
5

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work

is required to bring an additional electron from infinity to the origin?
Physics
1 answer:
aksik [14]3 years ago
5 0
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
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If the releases 39.4 kj of energy, how many kilocalories does it release? (1 cal = 4.184 j) (round off answer to 2 decimal place
REY [17]

By using unit conversion, the energy released in kilocalories is 9.42 kcal.

We need to know about unit conversion to solve this problem. There are several energy units which able to explain how much the energy is such as calories and joule. The energy unit can be converted to another unit by unit conversion. The unit conversion of calorie to joule is

1 cal = 4.184 joule

From the question above, we know that

E = 39.4 kJ

By using the unit conversion, we can convert the energy into calorie

E = 39.4 kJ

E = 39.4 x 10³ J

E = 39.4 x 10³ / 4.184 cal

E = 9.42 x 10³ cal

E = 9.42 x 10³ /10³  kcal

E = 9.42 kcal

Hence, the energy released in kilocalories is 9.42 kcal.

Find more on unit conversion at: brainly.com/question/141163

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8 0
2 years ago
a toy dart gun has a spring with k=128 N/m. a kid pulls back on the spring with a 8.22 N force. how far does it stretch?(unit=m)
german

Answer:

0.064 m

Explanation:

When a spring is stretched/compressed, the force exerted in the spring is related to the elongation of the spring by the equation:

F=kx

where:

F is the force applied

k is the spring constant

x is the stretching/compression of the spring

In this problem, we have:

F=8.22 N is the force applied by the kid on the spring

k = 128 N/m is the spring constant

Solving for x, we find how far the spring stretches:

x=\frac{F}{k}=\frac{8.22}{128}=0.064 m

8 0
3 years ago
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lina2011 [118]
*\\
From\ Newton's\ third\ Law:\\\\
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mass=5kg\\\\
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