Answer:




Explanation:
r = Radius of disk = 7.9 cm
N = Number of revolution per minute = 1190 rev/minute
Angular speed is given by

The angular speed is 
r = 2.98 cm
Tangential speed is given by

Tangential speed at the required point is 
Radial acceleration is given by

The radial acceleration is
.
t = Time = 2.06 s
Distance traveled is given by

The total distance a point on the rim moves in the required time is
.
Answer:
longitudinal and transverse.
Explanation:
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Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.
Answer:
1.30
Explanation:
To calculate the critical angle we have ti use the formula:

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.
By taking n_2 and replacing we obtain:

hope this helps!!