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Alex777 [14]
2 years ago
10

A proton accelerates from rest in a uniform electric field of 630 N/C. At one later moment, its speed is 1.50 Mm/s (nonrelativis

tic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Physics
1 answer:
Vlad1618 [11]2 years ago
8 0

Answer:

the acceleration of the proton is 6.025 x 10¹⁰ m/s².

Explanation:

Given;

magnitude of electric field, E = 630 N/C

final speed of the proton, v = 1.5 M m/s = 1.5 x 10⁶ m/s

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

The force experienced by the proton is calculated as;

F = ma = EQ\\\\a = \frac{EQ}{m} \\\\a = \frac{(630)(1.6\times 10^{-19})}{1.673 \times 10^{-27}} \\\\a = 6.025 \times 10^{10} \ m/s^2

Therefore, the acceleration of the proton is 6.025 x 10¹⁰ m/s².

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W=225 J<br> P=25 W<br> t=?<br> what's T
Mama L [17]

Explanation:

p=w/t

t=w/p

t=225/25=9seconds

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If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.
Alex Ar [27]
The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

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3 years ago
27 degrees to kelvin, 20 dg to milligrams, and 3 to decimeters.
Gala2k [10]
K = C + 273, so 27°C = 27+273 = 300 K

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3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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What did Charles Darwin’s conclude on the Galapagos Island? Plz answer fast
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He discovered several species of finches that varied from island to island and it helped him make his theory of natural selection.

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