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3 years ago

~cosmic snowball

Physics

2 answers:

gizmo_the_mogwai [7]3 years ago

5 0

The correct answer is D meteoroid

PIT_PIT [208]3 years ago

3 0

Answer:

Hi low ok kyuh a asterion and coment

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Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba

NISA [10]

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

-1786.5J

7 0

3 years ago

A boat takes 3.0 hours to travel 50 km down a river, then 5.4 hours to return. Determine the speed of the water in the river.

Nutka1998 [239]

Answer:

3.7 km/h

Explanation:

Let's call v the proper speed of the boat and v' the speed of the water in the river.

When the boat travels in the direction of the current, the speed of the boat is:

v + v'

And it covers 50 km in 3 h, so we can write

$v+v' = \frac{50 km}{3 h}=16.7 km/h$ (1)

When the boat travels in the opposite direction, the speed of the boat is

v - v'

And it covers 50 km in 5.4 h, so

$v-v'=\frac{50 km}{5.4 h}=9.3 km/h$ (2)

So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

$v+v'=16.7 \\v-v'=9.3$

Subtracting the second equation from the first one we get:

$(v+v')-(v-v')=16.7-9.3\\2v'=7.4\\v'=3.7$

So, the speed of the water is 3.7 km/h.

5 0

4 years ago

How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

irinina [24]

Answer:

Explanation:

Work = Force times displacement. Therefore,

W = 3150(75.5) so

W = 238000 N*m

6 0

3 years ago

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):