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zalisa [80]
3 years ago
15

A source is more reliable if _____.

Physics
1 answer:
AlekseyPX3 years ago
4 0
A is the correct answer
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A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this
mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0
3 years ago
Why you do scientists collect multiple trials?
sergiy2304 [10]

"When we do experiments it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time."

6 0
3 years ago
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0
3 years ago
An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s
gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

Q = ms\Delta T + mL

here we have

Q = 40(950)(680 - 32) + 40(450 \times 10^3)

Q = 24624 kJ + 18000 kJ

Q = 42624 kJ

Since this is the amount of aluminium per hour

so power required to melt is given by

P = \frac{Q}{t}

P = \frac{42624}{3600} kW

P = 11.84 kW

Since the efficiency is 85% so actual power required will be

P = \frac{11.84}{0.85} = 13.93 kW

Part B)

Total energy consumed by the furnace for 30 hours

Energy = power \times time

Energy = 13.93 kW\times 30 h

Energy = 417.9 kWh

now the total cost of energy consumption is given as

R = P \times 20 \frac{Cents}{kWh}

R = 417.9 kWh\times  20 \frac{cents}{kWh}

R = 8357.6 Cents

3 0
3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

So due to wind time will be more

4 0
3 years ago
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