1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
klio [65]
3 years ago
9

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa

l force on one proton due to the other proton?
Physics
2 answers:
Strike441 [17]3 years ago
3 0

Answer:

1.24\times 10^{36}

Explanation:

q = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

m = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

F_{e} = \frac{kq^{2}}{r^{2}}

F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}

F_{e} = 230.4 N

Gravitational force between the two protons is given as

F_{g} = \frac{Gm^{2}}{r^{2}}

F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}

F_{g} = 1.86\times 10^{-34} N

Ratio is given as

Ratio =\frac{F_{e}}{F_{g}}

Ratio =\frac{230.4}{1.86\times 10^{-34}}

Ratio = 1.24\times 10^{36}

kolbaska11 [484]3 years ago
3 0

Answer:

F_e:F_g = 1:1.237 * 10^{36}

Explanation:

Parameters given:

Distance between the protons, r = 1 * 10^{-15} m

Electric charge of proton, Q = 1.602 * 10^{-19} C

Mass of proton, m = 1.673 *10^{-27} kg

The electric force on one proton due to the other given as:

F_e = \frac{k*Q*Q}{r^2}

where k = Coulomb's constant

F_e = \frac{9 * 10^9 * 1.602 * 10^{-19} * 1.602 * 10^{-19} }{(1 * 10^{-15})^2} \\\\\\F_e = 230.98 N

The gravitational force on one proton due to the other is given as:

F_g = \frac{G*m*m}{r^2}

where G = gravitational constant

F_g = \frac{6.67408 * 10^{-11}*1.673 *10^{-27} * 1.673 *10^{-27}}{(1 * 10^{-15})^2} \\\\\\F_g = 1.868 * 10^{-34} N

Therefore, the ratio of electric force, F_e, and gravitational force, F_g, is:

\frac{F_e}{F_g} = \frac{230.98}{1.868 * 10^{-34}} = \frac{1}{1.237 * 10^{36}} \\\\\\F_e:F_g = 1:1.237 * 10^{36}

You might be interested in
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

= 4.75 m/s

6 0
3 years ago
To avoid an accident, a driver steps on the brakes to stop a 1000kg car traveling at 65km/h. If the braking distance is 35m, how
Alenkinab [10]
To stop the car it would be 100m because if the car is going to 65km/h then it would still be 100km/h
8 0
3 years ago
If a book has a a mass of 2 kg, how much does the book weigh?
Anestetic [448]

Answer:

A. 20 N

Explanation:

the weight of book is 2×10= 20N

7 0
3 years ago
If the ocular lens magnifies a specimen 10x, and the objective lens used magnifies the specimen 35x. What is the total magnifica
Kay [80]

Answer:

350x^2

Explanation:

I'm assuming you're saying you're using them together, like stacking them on top of each other, so if that's the case then this is a simple multiplication problem, which can be written as 10x(35x). Solve it and you get 350x^2.

8 0
2 years ago
A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The
vovikov84 [41]

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let \omega is the angular frequency. It is given by :

\omega=\sqrt{\dfrac{k}{m}}

\omega=\sqrt{\dfrac{475}{2.5}}

\omega=13.78\ rad/s

(c) Let T is the period. It is given by :

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{13.78}

T = 0.45 s

(d) Frequency,

f=\dfrac{1}{T}

f=\dfrac{1}{0.45}

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

v_{max}=\omega\times A

v_{max}=13.78\times 5.5\times 10^{-2}

v_{max}=0.757\ m/s

The value of acceleration is given by :

a=\omega^2A

a=(13.78)^2\times 5.5\times 10^{-2}

a=10.44\ m/s^2

Hence, this is the required solution.

7 0
3 years ago
Other questions:
  • Which planets are mostly "made of" atmosphere?
    15·2 answers
  • What do significant figures in a measurement include___________________________________.
    7·1 answer
  • All charged objects create an electric field around them. What two factors determine the strength of two electric fields upon th
    7·1 answer
  • What does combustion mean?what clues help u detemine the meaning?
    6·2 answers
  • Given 1 inch ≡ 2.54 cm and 1 foot ≡
    14·1 answer
  • ONLY 2 DAYS LEFT AND 5 WEEKS SUMMER BREAK FOR ME WHY NOT TODAY BE THE LAST DAY-
    5·1 answer
  • A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of
    7·1 answer
  • What is the frequency of this wave<br> 1<br> 2<br> 3<br> 4
    10·1 answer
  • HELLO I NEED YOU HELP WITH THIS SCIENCE QUESTION NO LINKS!!!
    5·1 answer
  • Which of the following describes the renewability of utility-scale wind energy?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!