Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Given,
Radius of flywheel (r) = 1.96 cm = 0.0196m
Angular acceleration (α)= 13.0 rad/s²
The tangential acceleration formula is at=rα
where, α is the angular acceleration, and r is the radius of the circle.
using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².
The tangential acceleration is 0.2548 m/s².
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The velocity of the two balls after the collision is 0.73 m/s.
The velocity of the two balls after the collision can be calculated using the formula below.
<h3>Formula:</h3>
- mu+m'u' = V(m+m')............... Equation 1
<h3>Where:</h3>
- m = mass of the first ball
- m' = mass of the second ball
- u = initial velocity of the first ball
- u' = initial velocity of the second ball
- V = velocity of the two balls after the collision.
make V the subject of the equation
- V = (mu+m'u')/(m+m')................ Equation 2
From the question,
<h3>Given:</h3>
- m = 0.25 kg
- m' = 0.3 kg
- u = 1 m/s
- u' = 0.5 m/s
Substitute these values into equation 2
- V = [(0.25×1)+(0.3×0.5)](0.25+0.3)
- V = 0.4/0.55
- V = 0.73 m/s.
Hence, the velocity of the two balls after the collision is 0.73 m/s
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Research in the field has more uncontrollable variable than in the laboratory. In the lab, you can conduct research in a controlled environment.
<span>The transit or passage of a planet across the disk of the Sun may be thought of as a special kind of eclipse. As seen from Earth, only transits of the inner planets Mercury and Venus are possible. Planetary transits are far more rare than eclipses of the Sun by the Moon.</span>
Answer:
Explanation:
a ) density = mass / volume
density of mars / density of earth
= (mass of mars/mass of earth) x (volume of earth / volume of mars)
= (mass of mars/mass of earth) x (diameters of earth / diameters of mars)³
= 0.11 x (130/69)³
= .73
b )
gravitational acceleration (g) = GM/R²
Gravitational acceleration on mars / Gravitational acceleration on earth
= (mass of mars / mass of earth) x (radius of earth / radius of mars )²
= .11 x (130/69 )²
= .39
Gravitational acceleration on mars = .39 x Gravitational acceleration on earth
Gravitational acceleration on mars = 9.8 x .39
= 3.82 m/s
c ) Escape velocity
c ) escape speed of mars / escape velocity of earth
= [( g on mars / g on earth ) x ( radius of mars / radius of earth )]°⁵
= √(.39 x 69/130)
= .4549
escape speed of mars = .4549 x escape velocity of earth
= .4549 x 11.2km/s
= 5.1 km / s