Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
renewable, you can keep growing corn for it. more fossil fuel can be made by nature but it takes way too long to be considered renewable
Answer:8.968 N-m
Explanation:
Given
Length of arm=0.152 m
Downward force=118 N
angle made by arm with vertical
Force can be divided into two components
It's sin component will contribute towards torque while cos component will not contibute
T=8.968 N-m
By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is
<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²
where
<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant
• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet
Solve for <em>M</em> :
<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)
<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))
<em>M</em> ≈ 2.8 × 10¹⁴ kg
