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3 years ago

10

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

ase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.
What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase? Express your answer with the appropriate units.

1 answer:

SashulF [63]3 years ago

4 0

Answer:

116 N.

Explanation:

Given,

mass of the ball, m = 0.145 Kg

initial speed = 0 m/s

Final speed = 40 m/s

time = 50 ms = 0.05 s

Force is equal to change in momentum per unit time

$F = \dfrac{m\Delta v}{\Delta t}$

$F = \dfrac{0.145(40-0)}{0.05}$

F = 116 N

Force of the pitcher hand on the ball is equal to 116 N.

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The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many

PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information

number of half-lives past is

= 15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

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2 years ago

How much energy is transferred in 10 seconds with a current of 13 amperes and a potential difference of 230 volts?

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29900 J

Explanation:

Recall that

P = VI

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Also,

E = Pt

= (2990 watts)(10 s)

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3 years ago

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with $m_{Pa}$ and its initial velocity with $u_{Pa}$

Let represent the mass of Ricardo with $m_{Ri}$ and its initial velocity with $u_{Ri}$

At rest ;

their velocities will be zero, i.e

$u_{Pa}$ = $u_{Ri}$ = 0

The initial momentum for this process can be represented as :

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = 0

after push off from each other then their final velocity will be $v_{Pa}$ and $v_{Ri}$

The we can say their final momentum is:

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

Since the initial velocities are stating at rest then ; u = 0

$m_{Pa}$ (0) + $m_{Pa}$ (0) = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

$m_{Pa}$ $v_{Pa}$ = - $m_{Ri}$ $v_{Ri}$

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So $m_{Ri} > m_{Pa}$ ;

Then $\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}$

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

$m_{Pa}$ $v_{Pa}$ = $m_{Ri}$ $v_{Ri}$

The ratio is

$\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1$

$v_{Pa} >v_{Ri}$

Therefore, Paula will have a greater speed than Ricardo after the push-off.

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Explanation is in the file

tinyurl.com/wpazsebu

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3 years ago

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Digiron [165]

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

$p_1 = p_2$

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Here,

$m_{1,2} =$ Mass of each object

$u_{1,2} =$ Initial velocity of each object

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Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

$m_1u_1 = (m_1+m_2)v_f$

Rearranging to find the velocity,

$v_f = \frac{m_1u_1}{ (m_1+m_2)}$

Replacing we have that,

$v_f = \frac{(332)(2.3)}{ (332+19.5)}$

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3 years ago