Answer:
5702.88 J or 5.7mJ
Explanation:
Given that :
C 1 = 6.0-μF
C 2 = 4.0-μF
V 1 = 50V
V 2 = 34V
Note that : Q = CV
Q 1 = C1 * V1
Q 1 = 50×6 = 300μC
Q 2 = 34×4 = 136μC
Parallel connection = C 1 + C 2
= 6+4 = 10μC
V = Qt/C
Where Qt = Q1+Q2
V = Q1+Q2/C
V = 300+136/10
V = 437/10
V = 43.6volts
Uc1 = 1/2×C1V^2
= 1/2 × 6μF × 43.6^2
= 1/2 × 6μF × 1900.96
= 3μF × 1900.96volts
= 5702.88J
= 5702.88J/1000
= 5.7mJ
Answer:
The gravitational acceleration experienced was of 1.63m/s².
Explanation:
We know, from the kinematics equations of vertical motion that:
Solving for g, we get:
Since the final speed is zero, because Neil Armstrong came to a stop in his maximum height, we obtain:
Finally, we plug in the given values of the initial speed and the maximum height:
This means that the gravitational acceleration experienced by Neil Armstrong in the moon, was of 1.63m/s².
Answer:
7200N
Explanation:
Centripetal force is directly proportional to the product of the mass and the square of the velocity and inversely proportional to the radius given.
Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
sound travels through air at around 340 - 350 m/s.
to cover 52 meters takes it, say, (52/345) = about 0.15 second
This would be the same time-lag even if it were not a major-league game,
except that you might not be able to get a seat 52 meters from home plate
at a little league or t-ball game.
