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dybincka [34]
3 years ago
11

A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.40 m,

what is the average resistance force exerted on her by the water
Physics
1 answer:
babymother [125]3 years ago
8 0

Answer:

19372.29 N

Explanation:

Kinetic Energy of the man = Work done by the average resistive force of water.

mg(H+d) = F×d............................ Equation 1

where, m = mass of the woman, H = Height of the platform, d = depth of water reached, F = average resistive force exerted by water.

make F the subject of the equation,

F = mg(H+d)/d...................... Equation 2

Given: m = 65.0 kg, H = 10 m, d = 3.4 m, g = 9.8 m/s²

Substitute into equation 2

F = 65(9.8)(100+3.4)/3.4

F = 19372.29 N

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avanturin [10]

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Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).

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8 0
3 years ago
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r
sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
3 years ago
For a wire has a circular cross section with a radius of 1.23mm.
Mila [183]

Answer:

5.731\times 10^{-5}\ m/s

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Conduction electron density in copper = 8.49\times 10^{28}\ electrons/m^3

v_d = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s

The drift speed of the electrons is 5.731\times 10^{-5}\ m/s

v_d=\dfrac{I}{neA}

From the equation we can see the following

v_d\propto \dfrac{1}{n}

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

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