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Musya8 [376]
3 years ago
8

Please help if you can.

Physics
1 answer:
VMariaS [17]3 years ago
3 0

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = P * t

E = 600 * 4 = 2400 Wh

E = 2.4 kWh

now we have total power consumed in 1 year

E = 365 * 2.4 = 876 kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P_1 = 876 * 10 = 8760 cents = $87.60

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = P * t

E = 300 * 4 = 1200 Wh

E = 1.2 kWh

now we have total power consumed in 1 year

E = 365 * 1.2 = 438 kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P_1 = 438 * 10 = 4380 cents = $43.80

total money saved in 1 year

Savings = 87.6 - 43.8 = $43.80

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What are the answers please help
aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.)  10.31 cm

e.)  10.24 cm  

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.)  If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm  

3 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
The cooling effect inside a refrigerator is produced by
nlexa [21]
Actually, when the refrigeration liquid is vaporized, the cooling effect is produced
6 0
3 years ago
Measurement conversions [metric to metric] 568 cm = m​
Trava [24]

Answer:

5.68 meters

Explanation:

hope this helps!

7 0
3 years ago
How is an elliptical galaxy different from a spiral galaxy?
laiz [17]
An elliptical galaxy is a type of galaxy having an approximately ellipsoidal shape and a smooth, nearly featureless brightness profile.

Unlike flat spiral galaxies with organization and structure, they are more three-dimensional, without much structure, and their stars are in somewhat random orbits around the center.
6 0
2 years ago
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