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3 years ago

Physical science

Physics

1 answer:

attashe74 [19]3 years ago

3 0

Answer:

0.56 km/s

Explanation:

We will define a single system of units for measurement, for this case meters per second [m/s]. That is, we must convert the rest of units such as centimeters per second and kilometers per second to meters per second.

$560[\frac{cm}{s}]*(\frac{1m}{100cm} )=5.6[m/s]\\0.56[\frac{km}{s}]*(\frac{1000m}{1km} )=560[m/s]$

Therefore the speed of 0.56 [km/s] is the greatest of all

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As the ball is moving in air as well as we have to neglect the friction force on it

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So here correct answer must be

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3 years ago

Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh

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Answer: Place his feet parallel to the baseline prior to tossing the ball

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2 years ago

Viewed through a spectroscope, the spectral profile of a yellow street lamp has a narrow line in the yellow region of the visibl

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Answer:

discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source

Explanation:

Bulbs can emit light in several ways:

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2 years ago

An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03

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Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium $\alpha =23.6\times 10^{-6}/^{\circ}C$

We know that change in length is given by $\Delta L=L\alpha \Delta T$

So $0.03=35\times 23.6\times 10^{-6}\Delta T$

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3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively