Answer:
-0.00152 V
Explanation:
Parameters given:
Diameter of the loop = 11 cm = 0.11m
Rate of change of magnetic field, dB/dt = 0.16 T/s
Radius of the loop = 0.055m
The area of the loop will be:
A = pi * r²
A = 3.142 * 0.055²
A = 0.0095 m²
The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:
EMF = - N * A * dB/dt
In this case, there's only one loop, so N = 1.
Therefore:
EMF = -1 * 0.0095 * 0.16
EMF = -0.00152 V
The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.
The correct answer is B, widespread pollution. If you look closely, you can see that the other answers are not problems at all, but benefits! :)
Answer:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Explanation:
The center of mass of a system of particles (), measured in meters, is defined by this weighted average:
(1)
Where:
- Mass of the i-th particle, measured in kilograms.
- Location of the i-th particle with respect to origin, measured in meters.
If we know that , , , , and , then the coordinates of the third particle are:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N
Therefore, for the weight of the cadet on the exoplanet, W₁, we have;
The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.