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Vedmedyk [2.9K]
3 years ago
7

Defination coulomb's law.

Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

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A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit
Tom [10]

1.97 E-7 m should be right. I took this a while ago.

6 0
3 years ago
The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant (
GuDViN [60]

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

Substituting the of the values variables into the equation, we have

C = ε2πrL/d

C =  24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C =  9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C =  996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

6 0
3 years ago
Two blocks, joined by a string, have masses of 6.0 and 9.0 kg. They rest on a frictionless, horizontal surface. A second string,
Lynna [10]

Answer:

T= 27 N

Explanation:

Assuming that the string joining both masses is massless  and inextensible, both masses accelerate at the same rate.

So, we can treat to both masses as a single system, and apply Newton's 2nd Law to both masses.

In this way, we can get the value of the acceleration without taking into account the tension in the string, as it is an internal force (actually a action-reaction pair).

Newton's 2nd law is a vector equation, so we can decompose the forces along perpendicular axis in order to convert it in two algebraic equations.

We can choose one axis as parallel to the horizontal surface (we call it x-axis, being the positive direction the one of  the movement of the blocks due to the horizontal force applied to the 6.0 kg block), and the other, perpendicular to it, so it is vertical (we call y-axis, being the upward direction the positive one).

Taking into account the forces acting  on both masses, we can write both equations as follows:

Fy = N- (m₁+m₂)*g = 0 (as there is no movement in the vertical direction)

Fx = Fh = (m₁ + m₂) * a ⇒ 45 N = 15.0 kg * a

⇒ a = 45 N / 15.0 kg = 3 m/s²

Now, in order to get the value of the tension T, we can choose as our system, to any mass, and apply Newton's 2nd Law again.

If we choose to the mass of 6.0 kg, in the horizontal direction, there are two forces acting on it, in opposite directions: the  horizontal applied force of 45 N, and the tension in the string that join both masses.

The difference of both forces, must be equal to the mass (of this block only) times the acceleration, as follows:

F- T = m₂* a ⇒ 45 N - T = 6.0 kg * 3 m/s²

⇒ T = 45 N -18 N = 27 N

We could have arrived to the same result taking the 9.0 Kg as our system, as the only force acting in the horizontal direction is just the tension in the string that we are trying to find out, as follows:

F = m₁*a = 9.0 kg* 3 m/s² = 27 N

4 0
3 years ago
The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in
Phantasy [73]

Answer:

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2)v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2)v = m1v1 + m2v2

m2v2= (m1 + m2)v – m1v1

v2 = ( (m1 + m2)v – m1v1)/m2

v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94

v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

v2 = 27.3m/s

5 0
3 years ago
A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw
ZanzabumX [31]

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

W = 2N

so we have

m = \frac{W}{g}

m = 0.204 kg

now tension force at the top is given as

T_{top} = 7 N

T_{bottom} = 15 N

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

F_{net} = T + W

F_{net} = 7 + 2 = 9 N

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

8 0
3 years ago
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