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photoshop1234 [79]
3 years ago
6

According to Coulomb’s Law, the force between two charged objects is related to _____.

Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

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

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

5 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
Mike rides his motorcycle North at an average speed of 20 meters/second for 50 seconds,
tia_tia [17]
He rode his bike 0.4 miles because you divide 20 by 50
5 0
3 years ago
a solid weighs 20gf in air and 18 gf in water.Find the specific gravity of the solid. Please show your work.​
ikadub [295]

Answer: It is given that A body weighs 20gf in air and 18. 0gf in water. Hence, the answer X-3 = 7.

5 0
3 years ago
1. A 2m-long and 3m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfa
ExtremeBDS [4]

Answer:

864 KN

Explanation:

(Absolute pressure) = (Atmospheric pressure) + (Gauge Pressure)

Atmospheric pressure = 95 KPa = 95000 Pa

Gauge Pressure = ρgh

ρ = density of the fluid = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = depth below the fluid level that the object is at = 5 m

Gauge Pressure = 1000 × 9.8 × 5 = 49000 Pa

Absolute pressure = 95000 + 49000 = 144000 Pa.

Pressure = (Hydrostatic force)/(Area perpendicular to the force)

Hydrostatic force = (Pressure) × (Area perpendicular to the force)

Area perpendicular to the force = 2 × 3 = 6 m²

Hydrostatic force on the top of the plate = 144000 × 6 = 864000 N = 864 KN

Hope this Helps!!!

8 0
3 years ago
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