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marshall27 [118]
3 years ago
14

What is the maximum centripetal acceleration experienced by a person standing still on the surface of the Earth? Where must they

be located?
Physics
1 answer:
skelet666 [1.2K]3 years ago
6 0

Answer:

The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Explanation:

Physically speaking, the centripetal acceleration (a_{r}), measured in meters per square second, experienced by a person is defined by the following expression:

a_{r} = \omega^{2}\cdot r (1)

Where:

\omega - Angular speed of the Earth, measured in radians per second.

r - Distance perpendicular to the rotation axis, measured in meters.

Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.

In addition, the angular speed of the Earth can be calculated in terms of its period (T), measured in seconds:

\omega = \frac{2\pi}{T} (2)

If we know that r = 6.371\times 10^{6}\,m and T = 86400\,s, then the maximum centripetal acceleration experienced by a person is:

a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)

a_{r} = 0.0337\,\frac{m}{s^{2}}

The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

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A car and an SUV both hit a patch of ice while driving eastbound down the road. The 1200-kg car was moving at 5 m/s when the 260
Radda [10]

The final speed of the SUV given the data from the question is 9.79 m/s

<h3>Data obtained from the question</h3>
  • Mass of car (m₁) = 1200 Kg
  • Velocity of car (v₁) = 5 m/s
  • Mass of SUV (m₂) = 2600 Kg
  • Velocity of SUV (v₂) = 12 m/s
  • Final velocity (v) =?

<h3>How to determine the final velocity of SUV</h3>

From conservative of linear momentum,

momentum before impart = momentum after impart

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Thus, we shall determine the final velocity of the SUV by obtaining the velocity after impart. This can be obtained as follow:

v = (m₁v₁ + m₂v₂) / (m₁ + m₂)

v = [(1200 × 5) + (2600 × 12)] / (1200 + 2600)

v = [(6000 + 31200)] / 3800

v = 37200 / 3800

v = 9.79 m/s

Thus, the final velocity of the SUV is 9.79 m/s

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brainly.com/question/250648

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5 0
2 years ago
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
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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

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