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crimeas [40]
3 years ago
11

On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea

rth?
Physics
1 answer:
snow_tiger [21]3 years ago
4 0

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

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A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch.A 8.2-V battery is connected
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Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance  in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})

Put the value into the formula

I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})

I(t)=0.01925\ A

I(t) = 19.25\ mA

(B). We need to calculate the store energy in the inductor

Using formula of energy

E=\dfrac{1}{2}LI^2

Put the value into the formula

E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2

E=7.04\times10^{-6}\ J

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

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