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Alex787 [66]
3 years ago
7

How strong is the attractive force between a glass rod with a 0.700μC0.700μC charge and a silk cloth with a –0.600μC–0.600μC cha

rge, which are 12.0 cm apart, using the approximation that they act like point charges?
Physics
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

0.2625 N/C

Explanation:

Force of attraction = \frac{K\timesQ_1\times Q_2}{R^2}

where Q₁ and Q₂ are charges and R is distance between charges.K is a constant equal to 9 x 10⁹.

= \frac{9\times10^9\times.7\times10^{-6}\times.6\times10^{-6}}{(12\times10^{-2})^2}

= 0.2625 N/C

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Guyz... Its my first question in this app... Pls do answer​
Basile [38]

Answer:

1.414

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1.  So:

1 sin 45° = n sin 30°

n = sin 45° / sin 30°

n = 1.414

Round as needed.

6 0
3 years ago
A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire
soldi70 [24.7K]

Answer:

a)  v=4.0m/s

b)  B=2.958T

Explanation:

From the question we are told that:

Wire Length l=10cm=>0.10m

Resistance R=0.35

Force F=1.0N

Power P=4.0W

a)

Generally the equation for Power is mathematically given by

P=Fv

Therefore

v=\frac{P}{F}

v=\frac{4.0}{1.0}

v=4.0m/s

b)

Generally the equation for Magnetic Field is mathematically given by

B=\frac{\sqrt{PR}}{vl}

B=\frac{\sqrt{4*0.35}}{4*0.10}

B=2.958T

5 0
3 years ago
When light reaches a barrier or the edge of an opening it diffracts, which means that the light _________. A) bends. B) bounces
rusak2 [61]
A. it bends when light reaches an end of a barrier it will bend.
3 0
3 years ago
Read 2 more answers
A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i
Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0
3 years ago
3. Identify the structures of the skin by the labels A through G below
ioda
I need help to I have a big test and I need help so bad
4 0
3 years ago
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