Aspeeding motorist traveling 120 km/h passes a stationary

Ответ:

kaitlynmoore42

10.03.2022

Physics

a. It will take 24.6 s for the police officer to reach the speeder

b. The police speed will be 66.42 m / s at this time

Further explanation

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

$\large{\boxed{\bold{S=v\times\:t}}}$

S = distance = m

v = speed = m / s

t = time = seconds

Straight motion changes regularly are the straight motion of objects that have a fixed acceleration

Formula used

$\large{\boxed{\bold{St=vot+\frac{1}{2}at^2}}}$

V = vo + at

Vt² = vo² + 2as

St = distance on t

vo = initial speed

vt = speed on t

a = acceleration

The full version of the question might be like this:

A speeding motorist traveling 120 km / h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.8 km / h / s (note the mixed units).

(a) How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed?

(b) How fast will the police officer be traveling at this time?

A speeding motorist traveling 120 km/h (fixed speed) shows irregular straight motion

Whereas The officer immediately begins pursuit at a constant acceleration of 10.8 km/h/s showing that straight motion changes irregularly

So that the officer can catch A speeding motorist, the distance achieved by both of them is the same and also with the same time

1. the distance achieved by the speeding motorist

v = 120 km / h = 33.3 m / s

S1 = v. t

S1 = 33.3t

2. the distance reached by the officer

vo = 0

a = 10 km / h / s = 2.7 m / s²

S2 = vot + 1/2 at²

S2 = 0 + 1/2 .2.7t²

S2 = 1.35t²

a. The distance traveled when the two meets are the same, so the time of both can be determined

S1 = S2

33.3t = 1.35t²

t = 24.6 s

b. The officer speed

Vt = vo + at

Vt = 0 + 2.7.24.6

Vt = 66.42 m / s

Learn more

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Keywords: the officer, a speeding motorist

, fixed acceleration, fixed speed

Aspeeding motorist traveling 120 km/h passes a stationary police officer. the officer immediately be

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Ответ:

desiree98

10.03.2022

Physics

120 km/h = 33.33 m/s 10.8 km/h/s = 3 m/s/s the motorist will cover a distance of 33.33 m x T(seconds) the police officer will cover a distance of 1/2 (3) T^2 they will be at the same point when 33.33T = 1.5 T^2 33.33 = 1.5 T 3T = 66.66 T = 22.22 seconds it will take the officer 22.22 seconds to catch the motorist. the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s almost 240 km/h (239.976 km/h)

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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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Aspeeding motorist traveling 120 km/h passes a stationary police officer. the officer immediately begins pursuit at a constant acceleration of 10.8

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