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zzz [600]
2 years ago
8

A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

the total change in energy of the system
Physics
1 answer:
AveGali [126]2 years ago
3 0

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

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dangina [55]

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

4 0
3 years ago
1. A rocket is fired vertically from the launch pad with a
zavuch27 [327]

Answer:

Max height= 36000 meters

Total Time = 120 seconds

Explanation:

0 = U - at

U = at

U= 20*60

U= 1200 m/s

MAX altitude would be

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Time of FLIGHT

2 * 1200/20

2400/20

120 sec onds

7 0
3 years ago
A wrench is accidentally dropped at the top of an elevator shaft in a tall building. (a) How many meters does the wrench fall in
RUDIKE [14]

Answer:

11.25m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}  \\  = 0 \times 1.5 +  \frac{1}{2} \times 10 \times   {1.5}^{2}  \\  = 5 \times 2.25 \\  = 11.25

6 0
3 years ago
Would anyone be kind enough to help me? :)
Travka [436]

Answer:

The racetrack is 996.7 meters long

Explanation:

Convert 251km/h to km/s (kilometers per second)

3600 seconds in an hour, so:

251/3600 = 0.0697km/s

Convert km/s to m/s (meters per second)

1000 meters in a kilometer, so:

0.0697*1000 = 69.7m/s

Find length of racetrack:

69.7m/s*14.3s = 996.7m

If the racer travels 69.7 meters in one second and it takes 14.3 seconds to complete a lap, the racetrack is 996.7 meters long.

6 0
3 years ago
A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t
Arisa [49]

A) 2.4\cdot 10^{-16}kg

The radius of the oil droplet is half of its diameter:

r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m

Assuming the droplet is spherical, its volume is given by

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3

The density of the droplet is

\rho=885 kg/m^3

Therefore, the mass of the droplet is equal to the product between volume and density:

m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg

B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

3 0
3 years ago
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