Answer: The electric field decreases because of the insertion of the Teflon.
Explanation:
If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.
However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.
In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
Answer:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN
Answer:
Distributes a floor load or weight
Explanation:
