Help please 20 points ? ASAP

when charge 2 is 3.0 m away from charge 1, the strength of the electric force on charge 2 by charge 1 is 0.80 n. if instead, cha

Nikolay [14]

The strength of the electrostatic force is inversely proportional to the square of the distance

between the charges. If the distance is doubled, the force is 1/4.

The new force is 0.80/4 = 0.20 N.

<h3>What is electrostatic force? </h3>

Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.

The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

$F=K |\frac{q_{1} q_{2}}{r^{2} } |$

In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:

Therefore, with q₁ and q₂ values of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = $9 \times 10^9$ N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is $8.854\times10^{-12} C^{2} N^{-1} m^{-2}$ .

To learn more about electrostatic force , visit:

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5 0

2 years ago

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W

vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

$f = 1.92 Hz$

Part c)

$speed = 3.65 m/s$

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

$T = 4\times 0.13 = 0.52 s$

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

$f = \frac{1}{T}$

$f = \frac{1}{0.52}$

$f = 1.92 Hz$

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

$speed = wavelength \times frequency$

$speed = 1.92 \times 1.9$

$speed = 3.65 m/s$

4 0

3 years ago

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

$a_c = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

$a_c = \frac{(12\ m/s)^2}{2\ m}$

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Now, the centripetal force applied by the athlete on the hammer will be:

$F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)$

6 0

2 years ago

All of the waves in the electromagnetic spectrum are _______ waves.

crimeas [40]

<span>All of the waves in the electromagnetic spectrum are transverse waves.</span>

5 0

3 years ago

Read 2 more answers

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu

erica [24]

Answer:

a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

Em_f = K = ½ m v²

as they indicate that there is no friction

Em₀ = em_f

mgh = ½ m v²

v = $\sqrt{2gh}$

let's calculate

v = $\sqrt{2 \ 9.8 \ 14.0}$

v = 16.57 m / s

b) the centripetal acceleration has the formula

a = v² / r

a = 16.57² / 14.0

a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

N-W = m a

N - mg = m ar

N = m (g + a)

let's calculate

N = 60.0 (9.8 + 19.6)

N = 1.76 10³ N

the relationship with weight is

N / W = 1.76 10³/( 60 9.8)

N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0

2 years ago