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2 years ago

27. A 0.150 kg baseball is thrown upward with an initial

Physics

1 answer:

Tema [17]2 years ago

4 0

Answer:

a. $1.47N$

b. $1.47N$

Explanation:

Given:

Mass of the baseball = $0.150kg$

The initial velocity of the baseball = 20m/s

The force acting on the baseball is only due to gravity and the air resistance is negligible. The baseball acceleration is constant due to gravity.

According to Newton's second law:

$F_{w} =mg$ ----------(1)

where mass of the baseball $m = 0.150kgAnd acceleration due to gravity [tex]g = 9.8m/s^{2}$

Put all the value in equation 1.

$F_{w} =0.150\times -9.8$

$F_{w} =-1.47N$

Therefore, the force on the ball when it reaches half of its maximum height is $1.47N$ .

Similarly, the force acting on the ball when its peak $1.47N$ .

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A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

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Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

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Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

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(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

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2 years ago

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Aleonysh [2.5K]

I uploaded the answer to a file hosting. Here's link:

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2 years ago

A bicycle rider travels 50.0 km in 2.5 hours. What is the cyclists speed

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Hi,
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