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N76 [4]
3 years ago
13

In the nucleus of an atom, two protons are separated by a distance of 10^-13 m. What is the Coloumb force between them? The char

ge of a proton is + 1.602 x 10^-19 C.
Physics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

F=0.0231\ N

Explanation:

<u>Electrostatic Force </u>

When two point charges q_1 and q_2 are placed at a distance d, they exert a force to each other whose magnitude can be computed by the Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

Where k is the constant of proportionality

k=9.10^9\ Nw.m^2/c^2

The information provided is

d=10^{-13}\ m

q_1=q_2=+ 1.602 x 10^{-19}\ C

Computing the Coloumb force between them

\displaystyle F=\frac{9x10^9\ 1.602 x 10^{-19} \ 1.602 x 10^{-19} }{(10^{-13})^2}

\boxed{F=0.0231\ N}

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio
LUCKY_DIMON [66]

Answer:

29274.93096 m/s

2.73966\times 10^{33}\ J

-5.39323\times 10^{33}\ J

2.56249\times 10^{33}\ J

-5.21594\times 10^{33}

Explanation:

r_p = Distance at perihelion = 1.471\times 10^{11}\ m

r_a = Distance at aphelion = 1.521\times 10^{11}\ m

v_p = Velocity at perihelion = 3.027\times 10^{4}\ m/s

v_a = Velocity at aphelion

m = Mass of the Earth =  5.98 × 10²⁴ kg

M = Mass of Sun = 1.9889\times 10^{30}\ kg

Here, the angular momentum is conserved

L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J

Kinetic energy at perihelion is 2.73966\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times  10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}

Potential energy at perihelion is -5.39323\times 10^{33}\ J

K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J

Kinetic energy at aphelion is 2.56249\times 10^{33}\ J

Potential energy is given by

P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}

Potential energy at aphelion is -5.21594\times 10^{33}\ J

6 0
3 years ago
Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile
Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150  =  110 miles

Displacement along the 110 North freeway = 150 - 175   = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0
3 years ago
Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a
Levart [38]
He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
6 0
3 years ago
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A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
2 years ago
In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak
polet [3.4K]

How can we experimentally verify newton's laws?

7 0
3 years ago
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