Answer:
A. Ahmed has a greater tangential speed than Jacques.
D. Jacques and Ahmed have the same angular speed.
Explanation:
Kinematics of the merry-go-round
The tangential speed of the merry-go-round is calculated using the following formula:
v = ω*R
Where:
v is the tangential speed in meters/second (m/s)
ω is the angular speed in radians/second (rad/s)
R is the angular speed in meters (m)
Data
dA = RA : Ahmed distance to the axis of rotation
dJ = RJ : Jacques distance to the axis of rotation
Problem development
We apply the formula (1)
v = ω*R
vA= ω*RA : Ahmed tangential speed
vJ= ω*RJ : Jacques tangential speed
Ahmed is at a greater distance from the axis of rotation than Jacques, then,
RA ˃ RJ and Ahmed and Jacques have the same speed ω, then:
vA ˃ vJ
Complete Question
The complete question iws shown on the first uploaded image
Answer:
a

b

Explanation:
Now looking at the diagram let take that the magnetic field is moving in the x-axis
Now the magnetic force is mathematically represented as
x B
Note (The x is showing cross product )
Note the force(y-axis) is perpendicular to the field direction (x-axis)
Now when the loop is swinging forward
The motion of the loop is from y to z to to x to y
Now since the force is perpendicular to the motion(velocity) of the loop
Hence the force would be from z to y and back to z
and from lenze law the induce current opposes the force so the direction will be from y to z to x
Now when the loop is swinging backward
The motion of the induced current will now be x to z to y
The efficiency of the first Carnot engine is
n1 = 1 - Th/T
The efficiency of the second Carnot engine is
n2 = 1 - T/Tc
The total efficiency of the engines put in series is
n = 1 - Th/Tc
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Answer:
114.86%
Explanation:
In both cases, there is a vertical force equal to the sprinter's weight:
Fy = mg
When running in a circle, there is an additional centripetal force:
Fx = mv²/r
The net force is found with Pythagorean theorem:
F² = Fx² + Fy²
F² = (mv²/r)² + (mg)²
F² = m² ((v²/r)² + g²)
F = m √((v²/r)² + g²)
Compared to just the vertical force:
F / Fy
m √((v²/r)² + g²) / mg
√((v²/r)² + g²) / g
Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:
√((12²/26)² + 9.8²) / 9.8
1.1486
The force is about 114.86% greater (round as needed).