Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem
Put the value into the formula
We need to calculate the magnitude of the charge q₃
Using formula of net force
Put the value into the formula
Hence, The value of charge q₃ is 40.46 μC.
Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Answer:
Gravitational force is <u>noncontact</u> force
Explanation:
Contact force occurs due to the contact between two different objects. Non-contact force occurs due to either attraction or repulsion between two objects such that there is no contact between these objects. There is no field linked with the contact force. ... Gravitational force is an example of a non-contact force.
<span>When the Sun’s energy moves through space, it reaches Earth’s atmosphere and finally the surface. This radiant solar energy warms the atmosphere and becomes heat energy. This heat energy is transferred throughout the planet’s systems in three ways: by radiation, conduction, and convection</span>
Answer:
288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.
Explanation:
If all other parameters are constant,
Electrostatic Force of attraction ∝ (1/r²)
F = (k/r²) = 72.0
If r₁ = r/2, what happens to F₁
F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units
