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1 year ago

Why is impulse and momentum important in sports like cricket??

Physics

1 answer:

labwork [276]1 year ago

7 0

<h3>IMPULSE AND MOMENTUM</h3>

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$\large \sf \underline{Question:}$

Why is impulse and momentum important in sports like cricket??

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$\large \sf \underline{Answer:}$

Impulse and momentum are important in sports like cricket because from the word "momentum" it describe itself like you're getting used of it or you know what to do like there's no stopping you.

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How are impulse and momentum related?

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What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz

deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

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2 years ago

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

Polar moment of Inertia

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

Maximum sustainable torque on the solid circular shaft

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

Maximum sustainable torque on the tubular shaft

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

Maximum sustainable power in the solid circular shaft

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

Maximum sustainable power in the tubular shaft

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

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2 years ago

Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through

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Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

= 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

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2 years ago

As air pressure decreases, what happens

Naya [18.7K]

Answer:

I believe it is B, not 100% sure though

Explanation:

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2 years ago

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

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Answer:

1.34352 kg

Explanation:

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h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

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2 years ago