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bixtya [17]
3 years ago
13

In a certain city, electricity costs $0.20 per kw·h. what is the annual cost for electricity to power a lamp-post for 8.00 hours

per day with (a) a 100.-watt incandescent light bulb (b) an energy efficient 25-watt fluorescent bulb that produces the same amount of light? assume 1 year = 365 days.
Physics
1 answer:
UNO [17]3 years ago
8 0

Part a)

per day electricity power consumed when 100 W bulb is used for 8 hours

P = 8 * 100 = 800 Wh

for one year consumption

E = 365 * 800 = 292 kWH

now the cost will be given

cost = 0.20 * 292 = $ 58.4

now when other energy efficient light is used

P = 8 * 25 = 200 Wh

for one year consumption

E = 365 * 200 = 73 kWH

now the cost will be given

cost = 0.20 * 73 = $ 14.6

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Balok diam diatas bidang miring pada sudut kemiringan 40° balok mulai bergerak,tentukan koefisien gesek statis antara balok dan
Pachacha [2.7K]

Answer:

0.84

Explanation:

m = Massa balok

g = Percepatan gravitasi

\theta = Sudut kemiringan

\mu = Koefisien gesekan statik antara balok dan bidang miring

Gaya balok karena beratnya diberikan oleh

F=mg\sin\theta

Gaya gesekan diberikan oleh

f=\mu mg\cos\theta

Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.

mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84

Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.

7 0
3 years ago
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

8 0
3 years ago
About how much longer can the Sun continue to generate energy by nuclear reactions in its core?
QveST [7]

Answer: 5billion years

Explanation: The sun produces energy through radioactive fusion reaction.

Nebula theory states that the gaseous particles of the Earth collapsed as a result of its own gravity which continuously lead to fusion reaction for the production of nuclear energy.

The Core of the Sun is that area up to 25% from the radius of the sun,here the pressure here range up to 250million atmosphere containing mainly hydrogen which gets converted in Helium molecule. The core is the center for energy production accounting for more than 98%, nuclear energy is transmitted at about 4.3million metric tons per second.

5 0
3 years ago
Do true blackbodies exist in nature?<br><br> Yes or No??
MaRussiya [10]

Answer:

no

Explanation:

4 0
3 years ago
Read 2 more answers
A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
Marrrta [24]

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003  m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

8 0
3 years ago
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