Complete Question
A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop is
a. counterclockwise
b. clockwise
c. there is no current in the loop because magnetic field is uniform
d. there is no current in the loop because magnetic field does not change
Answer:
Option A is the correct answer
Explanation:
According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.
To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.
Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.
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Answer:
44,640 ft
Explanation:
assuming the rocket started from rest, then v₀ = 0
2 min = 120 s
Δx = v₀t + 1/2at²
Δx = 0 + 1/2(6.2 ft/s²)(120 s)² = 44,640 ft ≈ 8.45 mi
Answer:
Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. It speeds through the vacuum of space at 186,400 miles (300,000 km) per second.
Explanation:
Hope this helps :))
The jogger would have ran about 10 km/h.
2. The earth is heated and its atmosphere is cooled by terrestrial radiation.
3. The earth is cooled and its atmosphere is heated by terrestrial radiation.
4. The earth is heated and its atmosphere is cooled by solar radiation.
Don't answer unless you know for sure. Thank you so much!
Answer: The option 4 is correct answer.
Explanation:
Terrestrial radiation is a long wave electromagnetic radiation. It originates from the earth and its atmosphere.
The sun emits a huge amount of energy. It travels across the space. The atmosphere is not directly heated by the solar radiation. It is heated by the terrestrial radiation that the planet itself emits.
When the land is heated then it emits radiation which heats up the atmosphere.
The earth is cooled and its atmosphere is heated by terrestrial radiation.
Therefore, the relationship between the earth, its atmosphere and radiation is correctly compared by statement 4.
Answer:
29.5 m/s
Explanation:
Volumetric flowrate = (average velocity of flow) × (cross sectional area)
Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s
Cross sectional Area of flow = πr²
Diameter = 0.00579 m,
Radius, r = d/2 = 0.002895 m
A = π(0.002895)² = 0.0000037629 m²
Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)
v = 0.000111/0.0000037629
v = 29.5 m/s
Given Information:
diameter of the nozzle = d = 5.79 mm = 0.00579 m
flow rate = 0.111 liters/sec
Required Information:
Velocity = v = ?
Answer:
Velocity = 4.21 m/s
Explanation:
As we know flow rate is given by
Flow rate = Velocity*Area of nozzle
Where
Area of nozzle = πr²
where
r = d/2
r = 0.00579/2
r = 0.002895 m
Area of nozzle = πr²
Area of nozzle = π(0.002895)²
Area of nozzle = 2.6329x10⁻⁵ m²
Velocity = Flow rate/area of nozzle
Divide the litters/s by 1000 to convert into m³/s
0.111/1000 = 1.11x10⁻⁴ m³/s
Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵
Velocity = 4.21 m/s
Therefore, the water exit the nozzle at a speed of 4.21 m/s