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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.
You might be interested in
- Find the velocity of a wave in the ocean if the wavelength is 2.0m and the frequency is 0.56Hz.
pentagon [3]
1.12 m/s is the velocity. You can get the velocity of a wave by multiplying the frequency and wavelength together. The product is the velocity.
6 0
3 years ago
Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the c
levacccp [35]

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

6 0
3 years ago
Need help with this physics question!
Fynjy0 [20]

Answer:

omg i need help with the same answer lol

Explanation:

i wish i can help but i need help on this hehe

6 0
3 years ago
A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t
Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

d = d_1 + d_2

given that

d_1 = 200 cos20 i + 200 sin20 j

d_1 = 187.94 i + 68.4 j

d_2 = -230 sin30 i + 230 cos30 j

d_2 = -115 i + 199.2 j

now the total distance is

d = (187.94 - 115)i + (199.2 + 68.4)j

d = 72.94 i + 267.6 j

now the magnitude of the distance is given as

d = \sqrt{72.94^2 + 267.6^2}

d = 277.4

also the direction is given as

\theta = tan^{-1}\frac{267.6}{72.94}

\theta = 74.7 degree

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0
3 years ago
Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

3 0
3 years ago
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