when a wave moves from a less dense medium to a denser one, most of the wave energy is refracted? A) true B) false
2 answers:
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Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.
B) The total distance the car travels is 200 meter during the 10 s period.
Explanation:
Given Data
Initial velocity of the car () = 20.0 m/s
Final velocity of the car () = 0 m/s
Time (in motion) =7.00 s
Time (in rest) =3 s
To find - A) car's deceleration while it came to a stop
B) the total distance the car travels in 10 s
A) The formula to find the deceleration is
Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)
Deceleration = (() - (
)) ÷ time (m/s²)
Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)
Deceleration = (- 20) ÷ 3 (m/s²)
Deceleration = - 6.6667 m/s²
(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)
The car's deceleration is - 6.6667 m/s² while it came to a stop
B) The formula to find the distance traveled by the car is
Distance traveled by the car is equals to the product of the speed and time
Distance = Speed × Time (meter)
Distance = 20.0 × 10
Distance = 200 meters
The total distance the car travels during the period of 10 s is 200 meters
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
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