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3 years ago

6

a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

ow much momentum does it have when it hits the catcher units

2 answers:

topjm [15]3 years ago

8 0

Answer:

p = 6.25 kg-m/s

Explanation:

It is given that,

Velocity of the baseball, v = 44.7 m/s

Weight of the baseball, W = 1.4 N

Firstly we need to find the mass of the baseball. It can be calculated as :

$m=\dfrac{W}{g}$

$m=\dfrac{1.4\ N}{9.8\ m/s^2}$

m = 0.14 kg

The momentum of an object is equal to the product of mass and velocity. It is given by :

$p=m\times v$

$p=0.14\ kg\times 44.7\ m/s$

p = 6.25 kg-m/s

So, the momentum it have when it hits the catcher is 6.25 kg-m/s. Hence, this is the required solution.

kumpel [21]3 years ago

3 0

1.4 N is a weight so calculating it's mass
1.4/9.8 = 0.1428 kg
momentum will be 0.1428*44.7 = 6.38 kgm/s

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2 years ago

Read 2 more answers

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$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

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$\omega_{o}$ - Initial angular speed, in revolutions per second.
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$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

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The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

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$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

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We kindly invite to check this question on rotational motion: brainly.com/question/23933120

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3 years ago

Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to

Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

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Vc = V x ( $\frac{Rc}{Rc + Ri}$ )

where

Rc = resistance of copper = $\frac{ρl}{a}$ (l = length , a = area, ρ = resistivity of copper)

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