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Veronika [31]
3 years ago
14

3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

nd acceleration of her swing. She has a device that measures the linear velocity of the bat during the swing. Her bat is 0.85 m long. On her first swing, the linear velocity of the bat just prior to the swing is 0 m/s, and 0.15 s later, just prior to ball contact is 16 m/s. What is the average angular acceleration of the end of her bat during the swing?
Physics
1 answer:
inessss [21]3 years ago
5 0

Answer:

The average angular acceleration is  \alpha =125.487 rad /s^2

Explanation:

From the question we are told that

  From the question we are told that

        The length of the bat is l = 0.85m  \

         The initial linear velocity is  u = 0 m/s

         The time is  t = 0.15s

         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

        Where w_f is the finial angular velocity which is mathematically evaluated as  

            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

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allsm [11]
1) The general equations of motion of the projectile on the x and y axis are:
x(t) = v_0 \cos \alpha t
y(t)=v_0 \sin \alpha t -  \frac{1}{2}gt^2
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First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of 15^{\circ}. To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
v_0 \sin \alpha t -  \frac{1}{2}gt^2 =0
t( v_0 \sin \alpha -  \frac{1}{2} gt)=0
that has two solutions: t=0 (beginning of the motion) and
t= \frac{2 v_0 \sin \alpha}{g}
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
x_1=v_0 \cos \alpha t = v_0 \cos \alpha ( \frac{2 v_0 \sin \alpha }{g} )= \frac{2 v_0^2}{g} \sin \alpha \cos \alpha
with \alpha=15^{\circ}.

If we call \beta the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
x_2 =  \frac{2 v_0^2}{g} \sin \beta \cos \beta
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
\sin \alpha \cos \alpha = \sin \beta \cos \beta
Now let's remind that \cos \theta= \sin (90^{\circ} -\theta) so that we can rewrite the equation as
\sin \alpha \sin (90^{\circ}-\alpha) = \sin \beta \sin (90^{\circ}-\beta)
and using \alpha=15^{\circ}:
\sin 15^{\circ} \sin (75^{\circ}) = \sin \beta \sin (90^{\circ}-\beta)
and we can see that there are two values of \beta that satisfy the equation: \beta=\alpha=15^{\circ} and \beta=75^{\circ}, which is the solution of our problem.

2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
7 0
3 years ago
The brainstem is to breathing and arousal as the limbic system is to memories and: (2 points)
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Answer:

The brainstem is to breathing and arousal as the limbic system is to memories and emotion. See the explanation below, please.

Explanation:

The limbic system develops certain responses to emotional stimuli (joy, anger, sadness, fear, pleasure). It interacts with the autonomic endocrine and nervous system. It is also related to sexuality, memory, attention, memories. It includes some areas such as hypothalamus, tonsil, hippocampus, among others.

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Answer: 0.091 m

Explanation:

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B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0
3 years ago
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pogonyaev
Hello,

The answer is option A "Venus".

Reason:

The planet Venus spins the wrong way many scientists are not sure why. Its not options B, C, or D because these planets spin the same way the as each other. (besides Venus) Therefore the answer is option A.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
5 0
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