Answer:
L = 1.11 x 
m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x 
m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x 
m
Length = ?
So,
we know that,
U = 1/2 C Δ
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ
52000 J = (0.5) x (C) x (
)
C = 0.28 F
And we also know that,
C = 
E = 8.85 x 
K = 3.7
A = 0.20 x L
d = 2.6 x m
Plugging in the values into the formula, we get:
0.28 = 
Solving for L, we get:
L = 1.11 x m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Answer:
The tube should be held vertically and perpendicular to the ground.
Explanation:
Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:
Reasoning:
The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.
Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.
So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.
hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.
His new number should be: 239.2 Liters
Answer: unless it's acted upon by an external force
Explanation: Newton first law of motion State that an object will continue in it state of rest or in motion, unless it is been acted upon by an external force
Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.
The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.
A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.
Learn more about beat frequency here : brainly.com/question/14157895
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