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olga55 [171]
3 years ago
8

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return

s to the juggler’s hand?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

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Answer:

NO

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Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

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     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

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          d = v_B * t_A

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Now this can also be represented as using equation of motion as

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Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

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aliya0001 [1]

To calculate the change in kinetic energy, you must know the force as a function of position. The work done by the force causes the kinetic energy change

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The work-energy theorem states that the change in kinetic enegy of an object is equal to the work done on the object:

\Delta E_k = W

where the work done is the integral of the force over the position of the object:

W=\int F(x) dx

As we see from the formula, the magnitude of the force F(x) can be dependent from the position of the object, therefore in order to solve correctly the integral and find the work done on the object, it is required to know the behaviour of the force as a function of the position, x.

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Answer:

CaO

Explanation:

mark me brainliest

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