ANSWER QUICK 30 POINTS

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

or

$D=6.75\times 10^5\ m$

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

or

$D=1.12\times 10^6\ m$

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

or

$D=1.68\times 10^6\ m$

Hence, this is the required solution.

7 0

3 years ago

During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, sl

xeze [42]

<span>During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, slowing down and increasing following distance. Adverse weather means that you are driving in difficult and dangerous conditions. Increasing following distance will help you to maintain safe driving and avoid tailgating. </span>

4 0

3 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

klio [65]

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

1)

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

2)

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

3)

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

4)

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$