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garik1379 [7]
3 years ago
9

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

the impact with the bat, how many impules of importance are used to find the final velocity of the bat
Physics
1 answer:
Ivanshal [37]3 years ago
8 0

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

(900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

36300 =  (900 x v_1) + (200 x v_2)

9v_1 + 2v_2 = 363 ..............(i)

9v_1 = 363 - 2v_2

v_1=\frac{363 - 2v_2}{9}

The mathematical expression for the conservation of kinetic energy is

\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

(363-2v_2)^2+18v_2^2=195129

(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

22v_2^2-145v_2-63360=0

Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

The negative velocity is neglected.

Therefore, substituting 96 m/s for v_2 in (i), we get

v_1=\frac{363-(2 \times 96)}{9}

     = 19

Thus, only impulse of importance is used to find final velocity.

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Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

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F=ma, so a=\frac{F}{m_{b}}.

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Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c
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Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _______ and the plate _______.
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Explanation:

Between a plate and the body of a bolt, the projected area is equal to the product of the bolt _Diameter of the bolt______ and the plate ___thickness____.

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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
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Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

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X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

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     m_B = M

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Let's find the tension

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Now we change the mass of the second block

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We seek tension for this case

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Let's look for the relationship between the tensions of the two cases

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   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
I select true helllpppp me
yanalaym [24]

Answer:

your right answer is true

hope it helps you

3 0
3 years ago
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