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Ivanshal [37]
2 years ago
6

What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

Physics
1 answer:
lutik1710 [3]2 years ago
4 0
It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.



Hopr it helps :)
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The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e
Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed (v), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi (1)

Where:

\Delta t - Rotation time, measured in seconds.

R - Radius of the Earth, measured in meters.

\phi - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that \Delta t = 86160\,s, R = 6.371\times 10^{6}\,m and \phi = 37.6819^{\circ}, then the tangential speed at Livermore is:

v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}

v\approx 284.001\,\frac{m}{s}

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0
3 years ago
A luge and its rider, with a total mass of 85 kg, emerge from adownhill track onto a horizontal straight track with an initial s
Verdich [7]

<span>(1)   </span>Through the Second Law of motion, the equation  for Force is:

                                 F = m x a

Where m is mass and a is acceleration (deceleration)

 

<span>(2)   </span>Distance is calculated through the equation,

                             D = Vi^2 / 2a

Where Vi is initial velocity

<span>(3)   </span>Work is calculated through the equation,

                          W = F x D

 

Substituting the known values,

Part A:

<span>(1)   </span>  F = (85 kg)(2 m/s^2) = 170 N

<span>(2)   </span>  D = (37 m/s)^2 / (2)(2 m/s^2)  = 9.25 m

<span>(3)   </span>  W = (170 N)(9.25 m) = 1572.5 J

 

Part B:

<span>(1)   </span> F = (85 kg)(4 m/s^2) = 340 N

<span>(2)   </span>D = (37 m/s)^2 / (2)(4 m/s^2) = 4.625 m

<span>(3)   </span><span> W = (340 N)(4.625 m) = 1572.5 J</span>

7 0
3 years ago
An experiment is conducted to determine how the arm length of a pendulum effects its period. The experimental design calls for v
Anestetic [448]

Okay so don't quote me on this but I believe the answer is A) I'm saying this because B and C make no sense. and you can't change the mass of something without changing it totally.

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3 years ago
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A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV
Bogdan [553]

Answer:

r = 6.6 x 10³ m = 6600 m

Explanation:

The potential at a distance from a charged sphere can be given as follows:

V = \frac{kq}{r}\\\\r = \frac{kq}{V}

where,

r = distance = ?

k = Colomb Constant = 9 x 10⁹ Nm²/C²

q = charge on sphere = 3.3 C

V = potential = 4.5 MV = 4.5 x 10⁶ V

Therefore,

r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}

<u>r = 6.6 x 10³ m = 6600 m</u>

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3 years ago
A ball is projected upward at time t = 0.0 s from a point on a roof 90 m above the ground. The ball rises, then falls and strike
ASHA 777 [7]
The answer is -32 hope it helps
7 0
3 years ago
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