Answer:
Explanation:
Given that,
A light bulb has a resistance of 2.9ohms
R = 2.9 ohms
And a battery of 1.5V is applied
V = 1.5 V
We want to find the rate of energy transformed
First we need to know what rate of energy is
Rate of energy implies that we want to find power. Power is the rate at which work is done
P = Workdone / time
Then,
In electronic, the power dissipated by a resistor is given as
P = V² / R
P = 1.5² / 2.9
P = 0.7759 W
P ≈ 0.776 W
So, the rate at which electrical energy transformed in the lightbulb is 0.776 Watts
Answer:
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Explanation:
Total force required = Mass x Acceleration,
F = ma
Here we need to consider the system as combine, total mass need to be considered.
Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg
We need to accelerate the group of rocks from the road at 0.250 m/s²
That is acceleration, a = 0.250 m/s²
Force required, F = ma = 1744 x 0.25 = 436 N
Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N
Answer:
Milliliters to Ounces Conversions
some results rounded
mL - fl oz
200.00 6.7628
200.01 6.7631
200.02 6.7635
200.03 6.7638
200.04 6.7642
200.05 6.7645
200.06 6.7648
200.07 6.7652
200.08 6.7655
200.09 6.7658
200.10 6.7662
200.11 6.7665
200.12 6.7669
200.13 6.7672
200.14 6.7675
200.15 6.7679
200.16 6.7682
200.17 6.7686
200.18 6.7689
200.19 6.7692
200.20 6.7696
200.21 6.7699
200.22 6.7702
200.23 6.7706
200.24 6.7709
mL fl oz
200.25 6.7713
200.26 6.7716
200.27 6.7719
200.28 6.7723
200.29 6.7726
200.30 6.7729
200.31 6.7733
200.32 6.7736
200.33 6.7740
200.34 6.7743
200.35 6.7746
200.36 6.7750
200.37 6.7753
200.38 6.7757
200.39 6.7760
200.40 6.7763
200.41 6.7767
200.42 6.7770
200.43 6.7773
200.44 6.7777
200.45 6.7780
200.46 6.7784
200.47 6.7787
200.48 6.7790
200.49 6.7794
mL fl oz
200.50 6.7797
200.51 6.7800
200.52 6.7804
200.53 6.7807
200.54 6.7811
200.55 6.7814
200.56 6.7817
200.57 6.7821
200.58 6.7824
200.59 6.7828
200.60 6.7831
200.61 6.7834
200.62 6.7838
200.63 6.7841
200.64 6.7844
200.65 6.7848
200.66 6.7851
200.67 6.7855
200.68 6.7858
200.69 6.7861
200.70 6.7865
200.71 6.7868
200.72 6.7872
200.73 6.7875
200.74 6.7878
mL fl oz
200.75 6.7882
200.76 6.7885
200.77 6.7888
200.78 6.7892
200.79 6.7895
200.80 6.7899
200.81 6.7902
200.82 6.7905
200.83 6.7909
200.84 6.7912
200.85 6.7915
200.86 6.7919
200.87 6.7922
200.88 6.7926
200.89 6.7929
200.90 6.7932
200.91 6.7936
200.92 6.7939
200.93 6.7943
200.94 6.7946
200.95 6.7949
200.96 6.7953
200.97 6.7956
200.98 6.7959
200.99 6.7963
Explanation:
To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by
Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is
According to the data given we have to,
PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is
On the other hand,
The total change of entropy would be,
Since 
the heat engine is not reversible.
PART B)
Work done by heat engine is given by
Therefore the work in the system is 100000Btu
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