Information about Earthquakes would normally be shown on a map called "geologic hazards"
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:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N - =
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Well, its in the air, so the air is "upon" the ball. and when it comes down...you catch it, and throw it, and get someone out, and win the game, and just keep doing that, and boooommm you're and pro baseball player. Life is good