Answer:
q₃ = - 13.0935 μC
Explanation:
Given
q₁ = q₂ = +7.67 μC
We use the equation
V = Kq/r
We can apply it as follows
V₁ = K*q₁/r₁ = K*q₁/(√2*L)
V₂ = K*q₂/r₂ = K*q₂/L
V₃ = K*q₃/r₃ = K*q₃/L
Then
V₁ + V₂ + V₃ = 0
⇒ (K*q₁/(√2*L)) + (K*q₂/L) + (K*q₃/L) = 0
⇒ (K/L)*((q₁/√2) + q₂ + q₃) = 0
⇒ (q₁/√2) + q₂ + q₃ = 0
Since q₁ = q₂
⇒ (q₁)((1/√2) + 1) + q₃ = 0
⇒ q₃ = - (q₁)((1/√2) + 1) = +7.67 μC*(1.7071)
⇒ q₃ = - 13.0935 μC
Answer:
K. E = 1125 J or 1.125KJ
Explanation:
Mass (m) =10kg
Velocity (v) = 15m/s
Kinetic energy = ½ mv²
K. E = ½ × 10 × 15²
K. E = 5 × 225
K. E = 1125Joules
K. E = 1125 J or 1.125KJ
I hope this was helpful, please mark as brainliest
Answer:
When your mom walks in the room, lol. I don't see a "figure A"
:D
Answer:
A. The closest point in the Moon's orbit to Earth
Explanation:
The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.
Let's see instead the names of the other options:
B. The farthest point in the Moon's orbit to Earth --> this point is called apogee
C. The closest point in Earth's orbit of the Sun --> this point is called perihelion
D. The Sun's orbit that is closest to the Moon --> this point has no specific name
Answer:
4.7 GHz
Explanation:
Applying,
v = λf................. Equation 1
Where v = velocity of the radio wave, λ = wavelength, f = frequency
make f the subject of the equation
f = v/λ.............. Equation 2
Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s
From the question,
Given: λ = 0.0644 meters
Constant: v = <em>3.00 x 10⁸ m/s</em>
Substitute these values into equation 2
f = (3.00 x 10⁸)/0.0644
f = 4.66×10⁹ Hz
f = 4.7 GHz
