In order to calculate the electric field strength, we may use the formula:
E = kQ/d²
Where Q is the charge and d is the distance between the charge and the test charge. Substituting the values into the equation:
E = (9 x 10⁹)(8.7 x 10⁻⁹) / (3.5²)
E = 6.39 Newtons per coulomb
Therefore, the answer is 6.4 Newtons/coulomb
Answer:
the smallest angle from the antennas is <em>47.3°</em>
Explanation:
We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.
Therefore, the relation is:
λ = c /f
where
- c is the speed of light constant
- λ is the wavelength
- f is the frequency
Thus,
λ = (3 × 10⁸ m/s) / (3.4 MHz)
= (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)
= 88.235 m
Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as
θ = sin⁻¹(λ / d)
where
- d is the distance between the two radio antennas
Thus,
θ = sin⁻¹(88.235 / 120)
<em>θ = 47.3 °</em>
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Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is <em>47.3 °</em>.
Answer:
The work must be negative or positive depending on direction of motion of the object
Explanation:
This is because
W= force x distance
And = F x S cosစ
And this is positive when
Theta is less than π/2 and negative when theta is less than or equal to π/2
Answer:
10. m/s
Explanation:
Given:
y₀ = 3.1 m
y = 0 m
a = -9.8 m/s²
t = 2.4 s
Find: v₀
y = y₀ + v₀ t + ½ at²
0 m = 3.1 m + v₀ (2.4 s) + ½ (-9.8 m/s²) (2.4 s)²
v₀ = 10.47 m/s
Rounded to two significant figures, the student's initial velocity was 10. m/s.
Complete question:
The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10⁻³ m², and the spacing between the plates is 6 x 10⁻³ m. There is no dielectric between the plates. What is the charge on the capacitor?
Answer:
The charge on the capacitor is 1.77nC
Explanation:
Given;
magnitude of electric field between the plates, E = 100 kV/m
Area of each plate, A = 2 x 10⁻³ m²
Distance between the plates, d = 6 x 10⁻³ m
Charge on the capacitor is calculated as;
Q = CV
V = Ed
Therefore, the charge on the capacitor is 1.77nC
