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JulijaS [17]
2 years ago
11

two protons are released from rest when they are 0.750 nm apart (a)what is the maximum speed they will reach ? when does this sp

eed occur? (b)what is the maximum acceleration they will achieve?when does this acceleration occur?
Physics
1 answer:
Keith_Richards [23]2 years ago
8 0

Answer:

And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second

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dexar [7]

Answer: electric potential difference

Explanation:

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Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel
Delicious77 [7]

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

E=2\frac{kQ}{r^2}

Rearranging for Q

Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}

Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC

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3 years ago
At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt
Olin [163]

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3 years ago
3. Order the following lengths from shortest to longest.
horrorfan [7]

Answer: d < a < c < b

3.3 cm < 400 mm < 170 m < 22 km

Explanation:

Firstly, let's convert these length to meters m, in order to work with the same units:

a. 400 millimeters

400 mm \frac{1m}{1000 mm}=0.4 m

b. 22 kilometers

22 km \frac{1000m}{1 km}=22000 m

c. 170 meters

Here the lenght is already in meters

d. 3.3 centimeters

3.3 cm \frac{1m}{100 cm}=0.033 m

Now that we have all the lengths in meters, we can order them from shortest to longest:

0.033 m < 0.4 m < 170 m < 22000 m

or

d < a < c < b

7 0
4 years ago
An electron is accelerated from rest by a potential difference of (24.5 A) V for a distance of (4.50 B) cm. Determine the de Bro
DiKsa [7]

Answer:

206 pm

Explanation:

We are given that

Potential difference,\Delta V=24.5+A V

Distance,d=4.5+B cm

We have to determine the de Brogile wavelength of the electron.

A=11 and B=5

\Delta V=24.5+11=35.5 V

d=4.5+5=9.5 cm

Charge on electron,q =1.6\times 10^{-19} C

Mass of electron=m=9.1\times 10^{-31} kg

Speed of electron,v=\sqrt{\frac{2q\Delta V}{m}}

Using the formula

v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 35.5}{9.1\times 10^{-31}}

v=3.53\times 10^6 m/s

de Brogile wavelength, \lambda=\frac{h}{mv}

Where h=6.626\times 10^{-34}

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1 pm=10^{-12} m

\lambda=206 pm

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3 years ago
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