Answer:
where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have
since string makes 30 degree angle with the vertical so we will have
where E = electric field intensity
Light is refracted when it crosses the interface from air to glass in which it moves more slowly.
Since the light speed changes at the interface, the wave length of the light must change too. The wave length decreases as the light enter the medium and the light wave changes direction.
Answer:
Height as seen by the professor = 38.2 m
Explanation:
Angle of throw = θ = 69°
Velocity of throw = v
X component of velocity = v₁ = v cos 69 = 0.3584 v m/s
Vertical component of the velocity = v₂ = v sin 69 = 0.9336 v m/s
v₂ / v₁ = tan 69 = 2.605
v₂ = 2.605 v₁.
Professor sees as if the x component of velocity =0
v (as seen by professor) + v' = 0
=> v as seen by professor = -v' = -10.5 m/s
This shows that y component of the ball's velocity is 2.605 times its x component of velocity.
with respect to the professor, there is only y component of velocity.
v₂' =v₂ = 2.605 ( -10.5) = 27.4 m/s.
Height as seen by the professor = (27.4)² / 2(9.8) = 38.2 m
Answer:
For C1, Q = 1.6125×10⁻³ C
For C2, Q = 6.25×10⁻⁴ C
Explanation:
Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.
From the question,
Q = CV........................ Equation 1
Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.
For the first capacitor,
Q = C1V............. Equation 2
Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V
Substitute into equation 2
Q = (6.45×10⁻⁶ )(250)
Q = 1.6125×10⁻³ C.
For the the second capacitor,
Q = C2V............. Equation 3
Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V
Q = (2.5×10⁻⁶ )(250)
Q = 6.25×10⁻⁴ C
