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Bas_tet [7]
3 years ago
9

I need help! Please help Me!

Physics
1 answer:
vladimir2022 [97]3 years ago
5 0

Answer:

answer is DE I hope it will help you please follow me

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Most of the resistance of the human body comes from the skin, as the interior of the body contains aqueous solutions that are go
ruslelena [56]

Answer:

a) Resistivity=R.A/L

Here, R=500000ohm, L=1.6m, A=2pi* (0.14/2)*0.002 m2=0.00088.........we have converted quantities into SI units

Resistivity=R.A/L

=500000*0.00088/1.6 = 440/1.6= 275 ohm.meter

b) pskin/pinterior =275/4.8=57.3

For second part, correct option is (A), because cross section area is smaller.

4 0
3 years ago
Which of these is an experiment generally regarded as being first carried out by James Joule?
kipiarov [429]
The correct answer for this question is this one: "measuring the temperature increase of water from doing work stirring it." This experiment is generally regarded as being first carried out by James Joule is this one, <span>measuring the temperature increase of water from doing work stirring it.</span>
5 0
4 years ago
What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes​
Ilia_Sergeevich [38]
Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
4 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.
bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

4 0
3 years ago
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