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ElenaW [278]
3 years ago
6

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

electrical force, Fe, applied by q1 on q2?

Physics
2 answers:
qwelly [4]3 years ago
8 0
Theres the pic of answer

belka [17]3 years ago
5 0
We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values, 
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶)  / (3 * 10⁻²)²

F = 100 N  [ East of positive charge ]

Hope this helps!
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An object has an average acceleration of +6.18 m/s2 for 0.266 s . At the end of this time the object's velocity is +9.90 m/s .
Archy [21]

At the start of the 0.266 s, the object's speed was 8.26 m/s.

The question can only be talking about speed, not velocity.

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A golfer hits a golf ball upwards at an angle. We can ignore air resistance on the ball.
Natalka [10]

Answer:

Check the diagram from the photo

Explanation:

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The six statements below represent Newton's three laws of motion and Kepler's three laws of planetary motion. Match each stateme
mote1985 [20]

Answer:

1. Force = mass x acceleration - Newton

2. A planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out  equal areas in equal times - Kepler

3. For any force, there is an equal and opposite reaction force - Newton .

4. An object moves at constant velocity if there is no net force acting upon it - Newton

5. The orbit of each planet about the Sun is an ellipse with the Sun at one focus  - Kepler.

6. More distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2-a3 - Kepler.

Explanation:

The three laws of planetary motion formulated by Johannes Kepler or Kepler's laws of planetary motion:

  1. The first law states that the planets move around the Sun in an elliptical orbit with the Sun at one of the foci.
  2. The second law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time.
  3. The third law states that the square of the orbital period (p) of a planet is directly proportional to the cube of the mean distance (a) from the Sun (or semi-major axis of its orbit) i.e., p² is proportional to a³.

The three laws of motion formulated by Sir Isaac Newton or Newton's laws of motion:

  1. The first law, also known as the law of inertia states that an object at rest or moves at a constant velocity will remain at rest or keep moving at a constant velocity unless it is acted upon by a force.
  2. The second law states that the total force (F) applied on an object is directly related to the acceleration (a) of that object produced by the applied force and the mass (m) of the object, i.e., F = ma (assuming the mass m is constant).
  3. The third law, also known as the law of action and reaction states that when an object exerts a force on another object, then the latter exerts a force equal in magnitude and opposite in direction on the former object i.e., for every action, there is an equal and opposite reaction. The example includes the recoiling of a gun when it fires a bullet forward.
5 0
3 years ago
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m
Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

I = \frac{P}{4\pi r^2},

Where

P = Power

r = Radius

Replacing the values that we have,

I = \frac{60}{(4*\pi (0.7)^2)}

I = 9.75 Watt/m^2

The definition of intensity tells us that,

I = \frac{1}{2}\frac{B_o^2 c}{\mu}

Where,

B_0 =Magnetic field

\mu = Permeability constant

c = Speed velocity

Then replacing with our values we have,

9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}

Re-arrange to find the magnetic Field B_0

B_o = 2.86*10^{-7} T

Therefore the amplitude of the magnetic field of this light is B_o = 2.86*10^{-7} T

6 0
3 years ago
If an object is dropped from a height of meters, the velocity (in ) at impact is given by v= √2gh , where g=9.8m/sec^2 is the ac
Neporo4naja [7]

Answer:

A.) V = 14 m/s

B.) h = 36.6 m

Explanation:

Given the formula v = √2gh

where g = 9.8m/sec^2 is the acceleration due to gravity.

A.) Determine the impact velocity for an object dropped from a height of 10 m.

Substitute height h in the given formula

V = √2gh 

V = √2 × 9.8 × 10

V = √196

V = 14 m/s

b. Determine the height required for an object to have an impact velocity of 26.8 m/sec (~ 60 mph). Round to the nearest tenth of a meter.

Substitute the velocity in the given formula and make height h the subject of formula.

26.8 = √2 × 9.8 × h

Square both sides

718.24 = 19.6h

h = 718.24 / 19.6

h = 36.64 m

h = 36.6 m

5 0
3 years ago
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