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3 years ago

The coulombic force between two ions is reduced to ______ of its original strength when the distance between them is quadrupled.

Physics

1 answer:

gayaneshka [121]3 years ago

6 0

Answer:

Explanation:

According to the coulombs law, the force existing vetween the ions is expressed as;

F = kQq/r² .... 1

Q and q are the ions

r is the distance between the ions

If the distance between the ion is quadrupled, then;

F2 = kQq/(4r)²

F2 = kQq/16r² ... 2

Divide equation 2 by 1;

F2/F = kQq/16r² ÷ kQq/r²

F2/F = kQq/16r² × r²/kQq

F2/F = 1/16

F2 = 1/16 F

Therefore the coulombic force between two ions is reduced to<u> 1/16 </u>of its original strength when the distance between them is quadrupled.

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Magnetic fields exist

Effectus [21]

Answer:

Magnetic fields exist near a magnet, farther away from a magnet, and within a magnet.

So, the answer is D. All of the above.

Let me know if this helps!

6 0

3 years ago

Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete

harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

$v=\dfrac{d}{T}$

d is distance, $d=2\pi r$ and 1/T = f (frequency)

$v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s$

So, the average speed of the rubber stopper is 18.84 m/s.

4 0

3 years ago

What is a synovial joint? (ANATOMY)

S_A_V [24]

Answer:

From Wikipedia:

"A synovial joint, also known as diarthrosis, joins bones or cartilage with a fibrous joint capsule that is continuous with the periosteum of the joined bones, constitutes the outer boundary of a synovial cavity, and surrounds the bones' articulating surfaces. The synovial cavity/joint is filled with synovial fluid."

8 0

2 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

2 years ago