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3 years ago

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Latitude and longitude picture.

1 answer:

TiliK225 [7]3 years ago

7 0

Answer:

where? please comment it

Explanation:

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A number of compounds containing the heavier noble gases, and especially xenon, have been prepared. One of these is xenon hexafl

aalyn [17]

Answer:

The answer is "82.2 torr"

Explanation:

moles of Xe:

$= \frac{0.06}{131.293} \\\\ =0.00045699313 \ \mol$

moles of $F_2$ :

$= \frac{0.0274}{38} \\\\= 0.00072105263\ \ mol$

moles of produced $XeF_2:$

$= 0.00024$

moles of left $Xe$ :

$= 0.00021$

Calculate the Pressure:

$= \frac{(0.08206\times 0.00024 \times 293)}{(.1)} + \frac{(0.08206\times 0.00021 \times 293)}{(.1)} \\\\= 0.10819611 \ \ atm \\\\ = 0.10819611 \times 760 \\\\ = 82.2 \ \ torr$

5 0

3 years ago

Explain how Newton’s third law describes the motion of a rocket. How could this apply to the motion of a robot on ice?

kolbaska11 [484]

Answer:

According to Newton's third law of motion to every action, there is an equal and opposite reaction. When a rocket moves in free space it ejects hot gases. ... It is then the “reaction force”. As a result, the rocket is accelerated or propelled in the opposite direction.

Launching a rocket relies on Newton's Third Law of Motion. A rocket engine produces thrust through action and reaction. The engine produces hot exhaust gases which flow out of the back of the engine. In reaction, a thrusting force is produced in the opposite reaction.

5 0

3 years ago

A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

<em>Now, force normal to the inclined plane:</em>

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

<em>Frictional force:</em>

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

<em>The component of weight along the inclined plane:</em>

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

<em>Now the total force required along the inclination to move at the top of hill:</em>

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

<em>Hence the work done:</em>

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

<em>Now power:</em>

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

Which of these period three elements from the periodic table would you predict to be the MOST metallic and why? A) Al B) Si C) P

Sauron [17]

Sorry I don’t even know

6 0

3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

4 years ago