The bottom of the hockey stick is called a?

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

3 years ago

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A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

$T^2=\frac{4\pi ^2}{GM} a^2$

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

$M = 3.5M_{sun} = 6.96*10^{30} kg$

$a=4.2AU = 6.28*10^{11} meters$

$G = 6.67*10^{-11}m^3kg^{-1}s^{-2}$

We put this into Kepler's law and get:

$T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.$

This when converted to years is 4.6 years.

4 1

3 years ago

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Why do you think the outer planets have such extensive systems of rings and moons, while the inner planets do not?

Inessa05 [86]

Answer:

Because of immense gravity

Explanation:

The formation of the Solar system was a very dynamic process. A lot of matter was thrown towards the outer solar system which further formed the Gas giants: Jupiter, Saturn, Uranus, and Neptune. The size of these outer planets is huge so is their gravity.

Because of their huge gravity a lot of matter which was scattered in the outer solar system got attracted towards them. This matter is what make the rings of the outer planets. Also, because of immense gravity they captured larger bodies thus making them their Moons.

6 0

3 years ago

What component of a longitudinal sound wave is analogous to a trough of a transverse wave?

anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.

6 0

3 years ago

Two friends are carrying a crate of mass 200 kg up a flight of stairs. The crate has length 1.25 m and height 0.500 m, and its c

Marysya12 [62]

Ans; see attached file for calculation and answer

Explanation:

4 0

3 years ago