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lesya692 [45]
2 years ago
7

The measured value of mass M in an experiment is M = 0.743 ± 0.005kg. The error in 2M is

Physics
1 answer:
omeli [17]2 years ago
8 0

Given the measured value of mass M in the experiment the error in 2M is  ; dM

<u>Given data :</u>

Measured value of mass M = ( 0.743 ± 0.005 ) kg

<h3>The error in 2M in the measured value </h3>

The error in 2M in the measured value M = 0.743 ± 0.005 kg is dM while the error in M² will be 2dM

Hence we can conclude that the measured value of mass M in the experiment the error in 2M is  ; dM.

Learn more about error in 2M : brainly.com/question/6650225

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A family has a furnace that uses natural gas to heat their home. They are moving to an area that offers the option of using ener
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8 0
2 years ago
Use your knowledge of waves to explain why echoes occur. Use your explanation to devise a system to measure distances to objects
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3 years ago
When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c
e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = e^{kt}

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = e^{-0.01t}

V = V₀e^{-0.01t}

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀e^{-0.01t}

0.1V₀ = V₀e^{-0.01t}

0.1V₀/V₀ = e^{-0.01t}

0.1 = e^{-0.01t}

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0
3 years ago
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