To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is 
Thanks for sharing that information. After extensive calculation,
we can say with assurance that after some number of seconds,
a loud "crunch" is perceived by the souls aboard the ill-fated vessel.
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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the magnitude of charge=q=8.76 x 10⁻⁵C
Explanation:
the magnetic force Fm is given by
Fm= q V B sinθ
q= charge
v= velocity= 2.5 x 10⁴ m/s
B= magnetic field strength= 8.1 x 10⁻²T
Fm= magnetic force= 7.5 x 10⁻² N
θ=25°
so 7.5 x 10⁻² =q (2.5 x 10⁴ ) (8.1 x 10⁻²) sin25
q=8.76 x 10⁻⁵C
the answers going to be A
