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Mrrafil [7]
3 years ago
9

A sample of oxygen occupies 560. ml when the pressure is 800.00 mm hg. at constant temperature, what volume does the gas occupy

when the pressure decreases to 700.0 mm hg?
Physics
1 answer:
raketka [301]3 years ago
7 0
Answer:
final volume = 640 ml

Explanation:
To answer this question, we will use Boyle's law which states that:
"At constant temperature, the product of the pressure and the volume of a given mass of gas will be constant"
Thereore:
P1V1 = P2V2
800*560 = 700*V2
V2 = 640 ml

Hope this helps :)
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Scientists have designed solar cells to trap solar energy and convert it to _____ energy.
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Scientists have designed solar cells to trap solar energy and convert it to D electrical energy. Its D because converting something has to do with recharging something and electronics have to charge. I hope I helped. 
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3 years ago
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You do 20 J of work pushing a crate up a ramp. If the output work from the inclined plane is 11 J, then what is the efficiency o
fomenos

Answer:

55%

Explanation:

take efficiency=power output/power input multiply by 100%

5 0
3 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
lbvjy [14]
-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s 

                                                                           = <span> 3,600 meters .</span>
___________________________________

The formula that has all of this in it is the formula for 
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is 
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds. 

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !  

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
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5 0
3 years ago
Two children are balanced on a seesaw that has a mass of 18.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the
Mashcka [7]

Answer:

1.28 m

Explanation:

As shown in the diagram attached,

According to the principle of moment,

For a body at equilibrium,

Sum of clockwise moment = sum of anticlockwise moment.

Taking moment about the pivot,

W₁(1.6)+W(0.133) = W₂(x)............... Equation 1

Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.

But,

W = mg

Where g = 9.8 m/s², m = mass of the body

Therefore,

W₁ = 26×9.8 = 254.8 N,

Wₓ = 18×9.8 = 176.4 N

W₂ = 34.4×9.8 = 337.12 N

Substitute these values into equation 1

(254.8×1.6)+(176.4×0.133) = 337.12(x)

407.68+23.4612 = 337.12x

337.12x = 431.1412

x = 431.1412/337.12

x = 1.2789

x ≈ 1.28 m

7 0
2 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
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