A car is driven 13 mi east and then certain distance due north, ending up at the position 25 degrees north of east of its initia
1 answer:
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Answer:
a. 6.69m/s
b. y=4.48m
c. t=1.43secs
Explanation:
Data given, acceleration,a=35m/s^2
distance covered,d=64cm=0.64m,
a. to determine the speed, we use the equation of motion
initial velocity,u=0m/s
if we substitute values we arrive at
b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2
and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.
Hence we can write the equation above again
if we substitute values we have
c. the time it takes to arrive at 1.83m is obtain by using the equation below
if we insert the values, we solve for t , hence t=1.43secs
Answer:
The additional charge that will flow from the battery onto the positive plate is 0.924 nC
Explanation:
Given;
Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²
Separation distance, d = 3.00 mm = 3 x 10⁻³ m²
voltage of the capacitor, V = 58.0 V
dielectric constant, k = 4.60
Initial Capacitance of the capacitor before the addition of dielectric material:
Initial charge across the parallel plates:
Q₁ = CV
Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C
Capacitance of the capacitor after the addition of dielectric material:
Cequ. = Ck
Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F
Final charge across the parallel plates:
Q₂ = Cequ. x V
Q₂ = 20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C
Additional charge = Q₂ - Q₁
= 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C
= 9.239 x 10⁻¹⁰ C
= 0.924 nC
Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC