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4 years ago

Identify two types of simple machines in this compund machine.

Physics

2 answers:

sleet_krkn [62]4 years ago

4 0

the answer would be "screw and lever!", thats the answer because the middle is made into a rod of metal rapped in metal to screw into the cork, and the lever part is the two handles on the side to lift it out of the bottle.

Lemur [1.5K]4 years ago

4 0

The answer should be a screw and lever

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A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

6 0

3 years ago

Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?

UNO [17]

Answer:

1.55 m

Explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

$V =\frac{k*q}{r}$

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

$r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m$

$r2 =\frac{k*q}{V2} = \frac{9e9 N*m2/C2*1.7e-8C}{54V} = 2.83 m$

The distance between both points, is just the difference between the radius of both spheres, as follows:

r₂ - r₁ = 1.55 m

5 0

3 years ago

A concave mirror has a radius of curvature of 10 cm. Find the location an height of the image if the distance of the object is 8

puteri [66]

R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3

HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
=-20/8
=-2.5

7 0

3 years ago

What happens to light when it passes through a lens?

Novay_Z [31]

B. Light refracts as it passes through a lens.

The bending of a ray of light also occurs when light passes into and out of a glass lens. ... Because a convex lens can cause rays of light to converge, it can produce an image on a screen.

7 0

4 years ago

Read 2 more answers

Please help with 1 and 4

KIM [24]

Number 4 is : a light hour is a measure of time in space used to calculate the length of time between one object to another and one light hour equals 1.079 × 109 kilometers or 1.079e+9

5 0

3 years ago