Answer:
The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
Explanation:
Given;
Radius of circular loop, R = 3.00 cm = 0.03 m
Current in the loop, I = 12.0 A
Magnetic field at the center of circular loop is given as;
B = μ₀I / 2R
Where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
R is the radius of the circular loop
I is the current in the loop
Substitute the given values in the above equation and calculate the magnitude of the magnetic field;
B = (4π x 10⁻⁷ x 12)/ 0.03
B = 5.0272 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
Answer:
The frictional force 
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
<span>A. Comparison
</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison
NOT:
<span>B. Contrast
C. Chronological order
D. Cause and effect</span><span>
</span>
Answer:
The correct answer is All of the above.
Explanation:
Answer:
300 Nm ; 300 J
Explanation:
Given that:
Force (F) = 20 N
Distance (d) = 15 m
The kinetic energy (Workdone) = Force * Distance
Kinetic Energy = 20N * 15m
Kinetic Energy = 300Nm
K. E = 1/2
