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3 years ago

According to coulombs law, how is the electric force related to the distance between two charges?

Physics

1 answer:

lorasvet [3.4K]3 years ago

4 0

Answer:

Explanation:

From the formula of coulombs law F = Kq1q2/square of r, we can say the electric force is indirectly related to square of r

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At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 23 kn

Arturiano [62]

Answer:

30.66 knots

Explanation:

Distance of ship A from B at noon = 50 NM

$\frac{da}{dt}$ = Velocity of ship A = 22 knots = 22 NM/h

$\frac{db}{dt}$ = Velocity of ship B = 23 knots = 23 NM/h

Distance travelled by ship A from noon to 3 PM = 22×3 = 66 NM

a = Total distance travelled by ship A = 50+66 = 116 NM

b = Total distance travelled by ship B till 3 PM = 23×3 = 69 NM

c = Distance between Ship A and B at 3 PM = √(116²+69²) = 134.97 NM

a²+b² = c² (Pythagoras theorem)

Now differentiating with respect to time

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{116\times 22+69\times 23}{134.97}=\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=30.66\ NM/h$

∴ The velocity with which the distance is changing at 3 PM (3 hours later) is 30.66 knots

5 0

2 years ago

When a river overflows its banks, it deposits sediment over a broad, flat area of land on both sides of the river. This broad fl

valentina_108 [34]

The answer is letter D: a floodplain

6 0

3 years ago

A well-trained athlete can run 400m in 47s, what is the athlete’s velocity?

AnnZ [28]

Answer:

8.51 m/s

Explanation:

Velocity = Displacement/Time

Velocity = 400 m ÷ 47 s

Velocity = 8.51 m/s

7 0

2 years ago

A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is

Klio2033 [76]

Answer:

θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

sin θ₂ = n₁ / n₂ sin θ

let's calculate

sin = 1 / 1.3 sin 0.23

sin = 0.175

θ= 0.17528 rad

let's reduce to degrees

θ = 0.17528 rad (180ª / pi rad)

θ = 10.28º

6 0

3 years ago

light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur

VashaNatasha [74]

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

$t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}$

where;

n is refractive index of soap film = 1.33

$t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t = 225.56 *10^{-9} \ m\\\\t = 225.56 \ nm$

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

4 0

3 years ago