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lara [203]
3 years ago
8

A 75-kg human footprint is 0.05 m when the human is wearing boots. Suppose you want to walk on snow surfce that can at most supp

ort a pressure of 3 kPa. What should the minimal total snowshoe area be?
Physics
1 answer:
Sholpan [36]3 years ago
6 0

Answer:

area is  0.245 m²

Explanation:

The equation for pressure is

       P = F / A

Where P is the pressure, F the force and A the area where the force is applied.

With the data we can calculate the pressure applied by the feet on the human on the floor

        P1 = W / A = mg / A

        P1 = 75 9.8 / 0.05

        P1 = 14700 Pa

Now they tell us that the maxi maque pressure supports the snow is 3 kPa = 3000 Pa, the way to decrease the pressure is to increase the area on which force is applied.

        P2 = F / A2

 

 In this case we will calculate the minimum area, so we use the maximum pressure that supports the snow (P2 = 3000 Pa) and the force remains the weight of man

        A2 = F / P2

        A2= mg / P2

        A2 = 75 9.8 / 3000

       A2 = 0.245 m²

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How does electric force depend on the amount of charge and the distance between charges
Gennadij [26K]
<h2> F = k×\frac{q1q2}{r2}</h2>

Explanation:

  • The attractive or repulsive forces which act between any two charged species is an electric force.
  • The electric force depends on the distance between the charged species and the amount of charge which can be calculated by the formula given as follows  

    F = k×\frac{q1q2}{r2}

where, K is coulombs constant, which is equal to -                                  9 x10^9 Nm^2/C^2.

  • The unit for K is newtons square meters per square coulombs.
  • This is known as Coulomb's Law.
6 0
3 years ago
Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied
Karolina [17]

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

E = \frac{Kq}{r^2}

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = \sqrt{2}\times a

where, a be the side of the square

So, Electric field at B due to charge at A

E_{A}=\frac{Kq}{a^2}

E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

EA = 6531.32 N/C

Electric field at B due to charge at C

E_{C}=\frac{Kq}{a^2}

E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}

Ec = 6531.32 N/C

Electric field at B due to charge at D

E_{D}=\frac{Kq}{2a^2}

E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}

E=\sqrt{8840.5^{2}+8840.5^{2}}

E = 12500 N/C

Explanation:

8 0
4 years ago
Read 2 more answers
Two trains leave the station at the same time, one heading east and the other west. the eastbound train travels at 85 miles per
kakasveta [241]
<span>Since the trains area headed in completely opposite directions, the rate at which they gain distance from each other is simply equal to the sum of the magnitudes of their velocities, in this case 85 + 75 = 160 miles per hour. Therefore, the amount of time it will take for them to be 352 miles apart is 352/160 = 2.2 hours, or 2 hours and 12 minutes.</span>
3 0
3 years ago
How can the distance between the first bright band and the central band be increased in a double-slit experiment?
Strike441 [17]

Answer:

Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.

Explanation:

The distance y from the central band to the first bright band is given by

y = \dfrac{ \lambda D}{d}

where \lambda is the wavelength of light (or any particle), D is the distance to the screen, and d is the slit separation.

From this equation we see that, by increasing the wavelength \lambda , increasing the distance from the screen D, and decreasing the slit separation d, we increase the distance between the first bright band and the central band.

Therefore, the 2nd choice "<em>Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.</em>" is correct.

8 0
4 years ago
A 5kg box rests on a table. a 3 kg box rests on top of the 5 kg box. what is the normal force from the table?
Murrr4er [49]

==>  The total mass resting on the table is (5 kg + 3 kg) = 8 kg.

==>  The total weight of that mass is (8 kg) x (9.8 m/s) = 78.4 newtons

==>  The boxes are stacked.  So the table doesn't know if the weight on it is coming from one box, 2 boxes, 3 boxes, or 100 boxes in a stack.  The table only knows that there is a downward force of 78.4 newtons on it.

==>  The table stands in a Physics classroom, and it soaks up everything it hears there.  It knows that every action produces an equal and opposite reaction, and that forces always occur in pairs.  

Ever since the day it was only a pile of lumber out behind the hardware store in the rain, the table has known that in order to maintain the good  reputation of tables all over the world, it must resist the weight of anything placed upon it with an identical upward force.  This is the normal thing for all good tables to do, up to the ultimate structural limit of their materials and construction, and it is known as the "normal force".

So the table in your question provides a normal force of 78.4 newtons. (d)

7 0
3 years ago
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