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Art [367]
3 years ago
6

In the theory of plate tectonics, various segments of Earths crust, called plates, move toward and away from each other. In one

instance, the plate that consists of the Indian subcontinent drifted from southeastern Africe to its current position in asia, traveling at a speed of 15cm/y. This plate collided with Asia, forming the Himalayan mountain range in the process. Most of this formation occurred during the last 1.00 x 10^7 years, during which time the indian subcontinents motion has slowed to about 5 cm/y. What has been the acceleration in units of cm/y^2, of the indian subcontinent during this time period.
Physics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

-1.00 × 10⁻⁶ cm/y²

Explanation:

Using a = (v - u)/t where a = acceleration of the indian subcontinent, v = final speed of the indian subcontinent = 5 cm/y, u = initial speed of the indian subcontinent = 15 cm/y, and t = time of motion of the indian subcontinent = 1.00 × 10⁷ years.

Substituting the values of the variables, we find the acceleration of the indian subcontinent, a  = (v - u)/t = (5 cm/y - 15 cm/y)/1.00 × 10⁷ years = -10 cm/y ÷ 1.00 × 10⁷ years = -1.00 × 10⁻⁶ cm/y²

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Explanation:

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
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