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zzz [600]
3 years ago
11

Ethel throws a rock downward, and air resistance is negligible. Compared to a rock that is dropped, the acceleration of the thro

wn rock (after it is thrown) is
Physics
1 answer:
Scorpion4ik [409]3 years ago
3 0

The accelerations will be the same because the acceleration of each rock is caused only by gravity, which accelerates objects the same amount no matter what the velocity is.

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According to Einstein’s theory of relativity, how can energy be created?
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What is the period of a wave if the wavelength is 100 m and the speed is 200 m/s?
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Read 2 more answers
Help and show work thanks
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3 years ago
A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s
Kobotan [32]
Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span>(65+0.045)*2.5
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Then equate them together : x=2.48 m/s 
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6 0
4 years ago
Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain
Luden [163]

Answer:

P=1362\ W

t'=251.659\ s is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

Q_v=1.1282\times 10^{6}\ J

Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

P=1362\ W

Now the time required to heat to boiling point form initial temperature:

t'=\frac{Q_s}{P}

t'=\frac{342760}{1362}

t'=251.659\ s

6 0
4 years ago
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