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Yuliya22 [10]
3 years ago
14

a girl of mass 40 kg wearing high heel of radius 3mm. find the pressure exerted by her on the ground. [answer the question fast]

Physics
1 answer:
kakasveta [241]3 years ago
7 0
You should calculate 40 kg and the radius 3mm.
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Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat
aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R    ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0
3 years ago
A force of 30n is applied to a screwdriver to pry the lid off of a can of paint. The screwdriver applies 75n of force to the lid
gizmo_the_mogwai [7]
The mechanical advantage of the screwdriver that is being described above is equal to 75N. This means that for every 30N that is applied on the screwdriver, this simple machine would in turn apply 75N of force to the lid of the can. 
3 0
3 years ago
nan moved 18m to the right and then another 22m to the right if the motion takes 20 seconds what is nans velocity
IrinaVladis [17]
Step 1: list known info

distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds

Step 2 :solve for velocity

velocity= Δx÷time
v= 40/20= 2m/s

Answer: the velocity is 2 meters per a second (m/s)


7 0
3 years ago
Help not to sure with this one need help plz asap​
DiKsa [7]

A is the answer for the problem

4 0
3 years ago
Read 2 more answers
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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