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mestny [16]
3 years ago
7

why it is possible to estimate, to a good approximation, the energy derived from a sugar cube in the body,by doing laboratory ex

periments ?
Physics
1 answer:
yarga [219]3 years ago
6 0
The sugar cube experiment in the laboratory  gives us a good approximation of the amount of energy that can be derived from the sugar cube because the amount of energy is neither created or destroyed, it is just converted to another form. If the energy from the sugar cube is converted to other form in the lab then, it is possible that same amount of energy will be derived. 
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The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A singl
dybincka [34]

Answer:

The expression is Vmem = ΔV - I*Rax

Explanation:

According to the picture, if switch S is closed, the Cmem will be short. If Vcap = 0, the current flows through the capacitor only (not through Rmem), thus Vmem = 0 after closing S.

When C is fully charged, then we have:

Vmem = ΔV - I*Rax

8 0
4 years ago
Each time the heart beats,what does it do to the blood
Tom [10]
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6 0
4 years ago
What is meant by lateral inversion​
grandymaker [24]

Answer:

Explanation:

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4 0
3 years ago
Which of the following is an example of conduction?
Jet001 [13]

the answer was C because Heat is transferred by conduction when a hot object is in direct contact with a cold object. So, a metal spoon becoming warm when placed in a cup of hot tea is an example of conduction because the spoon gains heat by being in direct contact with the hot tea.

7 0
4 years ago
An adiabatic closed system is accelerated from 0 m/s to 34 m/s. Determine the specific energy change of this system, in kJ/kg.
prohojiy [21]

Answer:

Δe=0.578 kJ/kg

Explanation:

Given data

Velocity v₁=0 m/s

Velocity v₂=34 m/s

to find

Specific energy change Δe

Solution

The specific energy change is simply determined from change in velocity

Δe=(v₂²-v₁²)/2

Put the given values to find the specific energy change

=(\frac{(34)^{2} *10^{-3} }{2} )\\=0.578kJ/kg

Δe=0.578 kJ/kg

6 0
3 years ago
Read 2 more answers
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