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Free_Kalibri [48]
3 years ago
5

places with continental climates typically have _______ summers and ________ winters warm; cool hot; cool warm; cold hot; cold

Physics
1 answer:
ipn [44]3 years ago
8 0
Places with continental climates typically have hot summers and cold winters. As compared to places with mild climates, places with continental climates tend to experience the two extremes when it comes to the seasons. Summers can get very hot, and winters can get very cold. 
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A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi
Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, \mu = 0.19

Now,

We know that the circumference of the circle is given by:

C = 2\pi R

27 = 2\pi R

R = \frac{27}{2\pi } = 4.29\ in

Also, in case of rolling, we know that:

Angular velocity, \omega = \frac{v}{R}

v = \omega R

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

m\omega' = m\omega + I\omega

Moment of Inertia, I = \frac{2}{5}mR^{2} = mk^{2}

mv'R = mvR + mk^{2}\times \frac{v}{R}

v'R = vR + k^{2}\times \frac{v}{R}

276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}

1184.04 - 756.903 = 4.29v

v = 99.56 in/s

Now,

Friction force, f = \mu mg

Also, acceleration of the ball can be computed as:

\mu mg = - ma

a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}

Now, the distance, d covered by the ball before rolling without slipping:

v^{2} = v'^{2} + 2ad

99.56^{2} = 276^{2} + 2\times (- 73.357)d

99.56^{2} - 276^{2} = 2\times (- 73.357)d

d = 451.65 in = 37.53 ft

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