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Verizon [17]
3 years ago
8

A proton accelerates from rest in a uniform electric field of 595 N/C. At one later moment, its speed is 1.15 Mm/s (nonrelativis

tic because v is much less than the speed of light). (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?
Physics
1 answer:
Neko [114]3 years ago
5 0
At one later moment, its speed is 1. 20Mm/s (non relativistic because v is much less than the speed of light).
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Two identical ions are each missing two electrons, each ion has a charge of 2e. What is the magnitude of the force between the i
xz_007 [3.2K]

Answer:

|F| = 2.09 × 10⁻⁸ assuming that the two ions are point charges.

Explanation:

What's the charge on each ion?

The symbol e here stands for fundamental charge. Each electron carries a negative fundamental charge of -e. Each proton carry a positive fundamental charge of +e.

Molecules and atoms are neutral. They contain an equal number of electrons and protons. Remove one electron from a molecule or atom, and that particle will end up with more protons (which are positive) than electrons. That particle will carry a positive charge of +e become an ion (a cation to be precise.) Remove another electron and the ion will carry a charge of +2e.

For each ion q = +2 \;e = 2\times 1.60\times 10^{-19}\;\text{C} = 3.2\times 10^{-19}\;\text{C}.

What's the size of the electrostatic force between the two ions?

Consider Coulomb's Law for the electrostatic force F between two point charges:

\displaystyle F = -\frac{k\cdot q_1\cdot q_2}{r^{2}},

where

  • k is Coulomb's constant,
  • q_1 and q_2 are the charge on the two point charges, and
  • r is the separation between the two charges.

Make sure that all values are in SI units. Assume that the two ions are small enough that they act like point charges:

\displaystyle \begin{aligned}F &= -\frac{k\cdot q_1 \cdot q_2}{r^{2}}\\&=-\frac{8.99\times 10^{9}\cdot(3.2\times 10^{-19}) \cdot (3.2\times 10^{-19})}{(2.1\times 10^{-10})^{2}}\\ &= -2.09\times 10^{-8}\;\text{N}\end{aligned}.

The value of F is negative, meaning that the two charges will repel each other because they are both positive. The question is asking for the magnitude of this force. Thus drop the sign in front of F to obtain 2.09\times 10^{-8}\;\text{N}, which is the magnitude of F.

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3 years ago
If the accepted value is 343 m/s, what is the best description of tonya’s data?
Nezavi [6.7K]
This depends on what are the values of Tonya's data. We can say that it is precise if the all data points are almost located at one area and it can be near or far from the accepted value. We can say that it is accurate when the data are near or equal to the accepted value.
7 0
3 years ago
Read 2 more answers
Calculate the efficiency of the following appliances:
lubasha [3.4K]

Answer:

1. The efficiency of the radiator is 90 %

2. The efficiency of the torch is 65 %

3. The efficiency of the car is 40 %

4. The efficiency of the energy saver is 70 %

5. The efficiency of the speaker is 50 %

Explanation:

Efficiency = (Useful energy out ÷ Total energy in) × 100 J

1. Useful energy = 900 J

The total energy in = 1000 J

The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %

2. Useful energy = 65 J

The total energy in = 100 J

The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %

3. Useful energy = 4,000 J

The total energy in = 10,000 J

The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %

4. Useful energy = 700 J

The total energy in = 1,000 J

The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %

5. Useful energy = 50 J

The total energy in = 100 J

The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %

4 0
3 years ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
Tom [10]

Answer:

a) The acceleration of the jogger is 1.5 m/s²

b) the acceleration of the car is also 1.5 m/s²

c) Yes, the car travels 76 m farther than the jogger.

Explanation:

a) The acceleration of an object is the variation of its velocity over time:

a = final velocity - initial velocity / time

for the jogger:

a = 3.0 m/s - 0 m/s / 2.0 s = <u>1.5 m/s ²</u>

b) For the car:

a = 41.0 m/s - 38.0 m/s / 2.0 s = <u>1.5 m/s²</u>

c) Let´s see the position of the car after 2 seconds.

The equation for the position of an accelerated object moving in a straight line is:

x = x0 + v0* t +1/2 * a * t²

Where:

x = position of the car at time "t"

x0 = initial position

v0 = initial velocity

t = time

a = acceleration  

 Let´s consider x0 = 0 because the origin of the reference system is located where the car starts accelerating. Then:

x = 38,0 m/s * 2 s + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 79 m

In the same way, we can calculate the position of the jogger:

x = 0 m/s * t + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 3 m

<u>The car travels 79 m - 3 m = 76 m farther than the jogger</u>

4 0
3 years ago
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