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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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If the weight of a person is 500 newton what is his mass on the earth ?
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The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

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Triss [41]

Answer:

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(b) 182.45 N

Explanation:

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Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+6450} \right )^2

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

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(b) h = 33700 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+33700} \right )^2

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

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3 0
3 years ago
A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6
Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

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Substitute the given parameters into the formula and calculate the value of a

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304 = 6.6248a

a = 304/6.6248

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Hence the average acceleration during this run is 45.89m/s²

4 0
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