Physics: Aplayer holds two baseballs a height (h) above the ground. he throws one vertically upward at speed (vo) and the other one vertically downward at the same speed. determine the speed of each ball when it strikes the ground and the difference between their times of flight.

The emissivity of the radiation shield is Explanation:From the question we are told that The temperature of the first parallel plate is The temperature of the second parallel plate is The emissivity of first plate is The emissivity of first plate is Generally the total radiation heat that is been transferred without the shield is mathematically represented as Where is the Stefan-Boltzmann constant which has a value Substituting values ...

Aplayer holds two baseballs a height (h) above the

Ответ:

queenkimm26

11.03.2022

Physics

Assume that air resistance may be ignored.
All measurements are positive upward.
g = acceleration due to gravity.

Case A; The ball is thrown vertically upward.
The time, t₁, to reach maximum height is one half of the time of flight.
Because the vertical velocity is zero at maximum height,
V₀ - gt₁ = 0
t₁ = V₀/g.
The time of flight is
t = 2t₁ = (2V₀)/g. (1)
Let v = the vertical velocity with which the ball strikes the ground. Then
v² = V₀² + 2(-g)(-h)
v = √(V₀² + 2gh) (2)

Case B: The ball is thrown vertically downward.
The time of flight, t, is given by
-h = -V₀t + (-g)t²
gt² + V₀t - h = 0
t = 1/(2g)[-V₀ +/- √(V₀² + 4gh)]
Reject negative time
$t=- \frac{V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1+ \frac{4gh}{V_{0}^{2}}$ (3)

Let v= the speed with which the ball strikes the ground.
-v = -V₀ - gt
v = V₀ + gt
$v = V_{0}- \frac{V_{0}}{2} + \frac{V_{0}}{2}\sqrt{1+ \frac{4gh}{V_{0}^{2}}} \\ v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}} })$

The speed with which the ball strikes the ground is
v = V₀√[1 + (2gh)/V₀²], when the ball is thrown upward
$v = \frac{V_{0}}{2} (1+\sqrt{1+ \frac{4gh}{V_{0}^{2}}})$ when the ball is thrown downward.

The difference in time of flight is
$\Delta t = \frac{3V_{0}}{2g} + \frac{V_{0}}{2g} \sqrt{1 + \frac{4gh}{V_{0}^{2}} }$

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Ответ:

tiakimberly3

04.10.2020

Physics

The emissivity of the radiation shield is e_3 = 0.580

Explanation:

From the question we are told that

The temperature of the first parallel plate is T_1 = 650K

The temperature of the second parallel plate is T_2 = 400K

The emissivity of first plate is e_1 = 0.6

The emissivity of first plate is e_2 = 0.9

Generally the total radiation heat that is been transferred without the shield is mathematically represented as

$Q_1 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{\frac{1}{e1 } + \frac{1}{e_2} -1 }$

Where $\sigma$ is the Stefan-Boltzmann constant which has a value $5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}$

Substituting values

$Q_1 = \frac{ 5.67 *10^{-8} (650 ^4 - 400 ^4)}{\frac{1}{0.6 } + \frac{1}{0.9} -1 }$

$Q_1 = 4876.8 \ W/m^2$

From the question we are told the that using the radiation shield would reduce the radiation heat transfer by 15%

So the new heat transfer is

$Q_2 = \frac{15}{100} * Q_1$

So $Q_2 = \frac{15}{100} * 4876.8$

Q_2 = 731.52 W/m^2

Now this new radiation heat transfer can be mathematically represented as

$Q_2 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{ [\frac{1}{e_1 } + \frac{1}{e_2 } - 1 ] + n [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

Where e_3 the emissivity of the radiation shield and n is the number of radiation shield

Substituting values

$731.52 = \frac{5.67 *10^{-8} (650 ^4 - 400 ^4)}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$731.52 = \frac{1.4175*10^{-5}}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = \frac{1.4175*10^{-5}}{731.52}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8}$

$1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8} - [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ]$

$\frac{1}{e_3} + \frac{1}{e_3} = 1 .7222222416$

e_3 = 0.580

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Aplayer holds two baseballs a height (h) above the ground. he throws one vertically upward at speed (vo) and the other one vertically downward at the same speed. determine the speed of each ball when it strikes the ground and the difference between their times of flight.

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