Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
Water and baking soda can be used, too.
Gamma rays, X-rays, most ultraviolet rays, and some infrared are absorbed by the atmosphere but do not reach the Earth's surface
The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
Learn more about the magnetic field of a circular arc here:
brainly.com/question/15259752
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