Answer:
I think it is 1.67
Explanation:
I multiply the 45 and 27 cause it says rate of which means to multiply.
Answer:
The three laws of motion were first compiled by Isaac Newton in his Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687. Newton used them to explain and investigate the motion of many physical objects and systems
100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.
The range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x 
90° 0
180° - 375 π x
We have current carrying wire in a form of a circle placed in a uniform magnetic field.
We have to the range of potential energies of the wire-field system for different orientations of the circle.
<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>
The formula to calculate the magnetic potential energy is -
U = M.B = MB cos 
where -
M is the Dipole Moment.
B is the Magnetic Field Intensity.
According to the question, we have -
U = M.B = MB cos
We can write M = IA (I is current and A is cross sectional Area)
U = IAB cos
U = Iπ
B cos
For = 0° →
U(Max) = MB cos(0) = MB = IπB = 5 × π × 
× 3 × 
=
375 π x .
For = 90° →
U = MB cos (90) = 0
For = 180° →
U(Min) = MB cos(0) = - MB = - IπB = - 5 × π × × 3 × =
- 375 π x .
Hence, the range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x
90° 0
180° - 375 π x
To solve more questions on Magnetic potential energy, visit the link below-
brainly.com/question/13708277
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Answer:
1.23×10⁸ m
Explanation:
Acceleration due to gravity is:
a = GM / r²
where G is the universal gravitational constant,
M is the mass of the planet,
and r is the distance from the center of the planet to the object.
When the object is on the surface of the Earth, a = g and r = R.
g = GM / R²
When the object is at height i above the surface, a = 1/410 g and r = i + R.
1/410 g = GM / (i + R)²
Divide the first equation by the second:
g / (1/410 g) = (GM / R²) / (GM / (i + R)²)
410 = (i + R)² / R²
410 R² = (i + R)²
410 R² = i² + 2iR + R²
0 = i² + 2iR − 409R²
Solve with quadratic formula:
i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)
i = [ -2R ± √(1640R²) ] / 2
i = (-2R ± 2R√410) / 2
i = -R ± R√410
i = (-1 ± √410) R
Since i > 0:
i = (-1 + √410) R
R = 6.37×10⁶ m:
i ≈ 1.23×10⁸ m
