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3 years ago

If you doubled the load resistor in a Wheatstone bridge, the load current would not be half as much. Why not?

1 answer:

6 0

Doubling the size of a load resistor increases the load current. Increasing the load resistance, in turn the total circuit resistance is reduced. The load current would not be half as much since when you increase the size of load resistor then load current increases.

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The atmosphere protects us from _____ and _____.

sveticcg [70]

Answer: B. meteorites, UV rays

The earth's atmosphere plays an important role in shielding the space rocks and meteorites enters inside the earth due to collision with the celestial body. Due to high heat and pressure in the mesospheric layer of the atmosphere these meteorites burned up before reaching the biosphere or hydrosphere system of the earth. The UV rays are protected from entering into the earth atmosphere by the ozone layer present in the stratosphere of the atmosphere. Both meteorites and UV rays can negatively effect the human population. Therefore, the atmosphere is an agent which provides protection against them.

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3 years ago

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a 70 kg skydiver opens her parachute. The force due to air resistance is now 1200 N. what is the acceleration of the skydiver

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Around 50 I think so

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3 years ago

A plane flies from basecamp the Lake A, 280 km away in a direction of 20° north of

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3 years ago

What does an epidemiologist do? (write it in your own words)

Vinvika [58]

An epidemiologist is a doctor who counteracts mass infections (epidemics, pandemics), organizes the treatment and prevention of the spread of epidemics.

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3 years ago

A wall with a surface area of 10 m2 is 2.5 cm thick. The inner surface temperature of the wall is 4150C, but the outer surface i

klio [65]

Answer:

650.65 K or 377.5°C

Explanation:

Area = A = 10 m²

Thickness of wall = L = 2.5 cm = 2.5×10⁻² m

Inner surface temperature of wall = $T_i$ = 415°C = 688.15 K

Outer surface temperature of wall = $T_o$

Heat loss through the wall = 3 kW = 3×10³ W

Thermal conductivity of wall = k = 0.2 W/m K

Assumptions made here as follows

There is not heat generation in the wall itself
The heat conduction is one dimensional
Heat flow follows steady state
The material has same properties in all directions i.e., it is homogeneous.

Considering the above assumptions we use the following formula

$Q=\frac {T_i-T_o}{\frac{L}{kA}}\\\Rightarrow T_o=T_i-\frac {QL}{kA}\\\Rightarrow T_o=688.15-\frac {3\times 10^{3}\times 2.5\times 10^{-2}}{0.2\times 10}\\\Rightarrow T_o=650.65~K$

∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C

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3 years ago