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balu736 [363]
3 years ago
11

What two simple machines make up an axe?

Physics
1 answer:
serg [7]3 years ago
7 0
It would be A since the axe head acts as a wedge and the handle is a basic lever
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1. Identify Which of the following is not a factor of global climate change:
Agata [3.3K]

Answer:

B.movement of continents

3 0
3 years ago
Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
what will happen to the brightness of the other bulbs in a series circuit if any of the bulbs were to burn out?
Ivenika [448]

Answer:

not work.

Explanation:

if u are saying in a series circuit...

if 1 build burns out and theres other bulbs the circuit wont work anymore.

6 0
3 years ago
A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia
slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0
3 years ago
Describe what happens to chemical<br> bonds during a chemical reaction
dimulka [17.4K]

Explanation:

The bonds that keep molecules together break apart and form new bonds during chemical reactions, rearranging atoms into different substances. Each bond takes a distinct amount of energy to either break or form; the reaction does not take place without this energy, and the reactants stay as they were.

6 0
2 years ago
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