6 . A transformer draws 24 W of power from a 220 V power line, stepping the voltage down to 7.5 V. What is the equivalent resistance of the input circuit?7.A circular loop of wire is in a spatially uniform magnetic field as shown. The magnetic field is directed into the screen and is perpendicular to the loop. What is the direction of the induced current in the loop, or is no current induced?
Answers
Answer:
6. 2017 Ω, about 2000 Ω.
7. no current is induced unless the field is changing
Explanation:
6. The relationship between power, voltage, and resistance is ...
P = V²/R
Rearranging gives
R = V²/P = (220 volt)²/(24 watts) ≈ 2016.667 Ω
The closest answer choice is 2000 Ω.
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7.No current is induced in a stationary loop in a time-invariant magnetic field.
If something were changing, the direction of current induced would depend on whether the field is increasing or decreasing, and/or whether the loop is expanding or contracting.
Final answer:
The equivalent resistance of the input circuit is approximately 2028.33 ohms.
Explanation:
In this question, we are given the power input and the voltage input of a transformer, and we are asked to find the equivalent resistance of the input circuit. To find the equivalent resistance, we can use the formula:
Resistance = Voltage2 / Power
Using the given values of voltage (220 V) and power (24 W), we can substitute these values into the formula to find the equivalent resistance:
Resistance = (2202) / 24 = 2028.33 ohms
Therefore, the equivalent resistance of the input circuit is approximately 2028.33 ohms.
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Answer:
C. Heat flows from lemonade to the ice
Explanation:
Transfer of heat from the drink to the ice is what melts the ice and cools the drink.
Answers
Answer:
Speed of train = 60 km/hr
Speed of plane = 160 km/hr
Explanation:
We can construct the following equations from the given information:
Speed of train * time = 150 -Equation 1
Speed of plane * time = 400 -Equation 2
Speed of plane = (3 * Speed of train) - 20 -Equation 3
We can solve these three simultaneous equations for the three variables (speed of train , speed of plane, time).
Substituting within Eq. 1 and Eq. 2 we get:
Speed of plane * 150 = 400 * Speed of train -Equation 4
Finally, solving Eq. 4 and Eq. 3 simultaneously we get:
Speed of train = 60 km/hr
Speed of plane = 160 km/hr
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If the class spent a month teaching it, then that's about how long it takes. Sorry.
Convert 6.36 x 104 from scientific notation to standard decimal with the correct number of significant digits.
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Answer:
Explanation:
standard decimal form of 6.36*10^4 is
6,3600.0
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Given: Takeoff Speed (u) = 28 m/s
Final velocity (v) = 0.0 m/s
Acceleration (a) = 1.9 m/s²
To calculate: The minimum length of runway (s) =?
Solution: Apply 3rd kinematic equation of motion
v² = u² - 2as
Or, s = ( u² - v²) / 2 a
Or, s = ( 28² - 0²) / 2 × 1.9 m
Or, s = 206.3 m
Hence, the required minimum length of the runway will be 206.3 m
Final answer:
To determine the minimum length of the runway a Cessna 150 airplane would need to take off, you can use a kinematic equation. Plugging in a final velocity of 28 m/s, initial velocity of 0 m/s (since the plane starts from rest), and acceleration of 1.9 m/s/s gives an answer of approximately 207.11 meters.
Explanation:
The subject of your question is related to physics, specifically, kinematics, which studies the motion of objects. In your case, a Cessna 150 airplane needs to accelerate from rest to a speed of 28 m/s for takeoff, with an average acceleration of 1.9 m/s/s, and you're trying to find out the minimum length of the runway required. For this, we can use a kinematic equation.
The equation we can use is the following: v^2 = u^2 + 2as, where v is the final velocity (28 m/s), u is the initial velocity (0 m/s since the plane starts from rest), a is the acceleration (1.9 m/s/s), and s is the distance we want to find out.
When you plug in the known values into the equation, you will get: (28)^2 = (0)^2 + 2 * 1.9 * s. After rearranging the equation, you will have s = [(28)^2 - (0)^2] / 2*1.9 = 207.11 m. Therefore, the minimum length of the runway required for the Cessna 150 airplane to take off is approximately 207.11 meters, assuming constant acceleration and no other factors interfering.
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