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mixas84 [53]
3 years ago
8

A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investig

ate the motion in which the string moves in a cone with half-angle θ. What tangential speed, v , must the bob have so that it moves in a horizontal circle with the string always making an angle θ from the vertical?Express your answer in terms of some or all of the variables m, L, and θ, as well as the acceleration due to gravity g.v = Lsin(θ) / sqrt(Lcos(θ) / g)v = Lsin(θ) / sqrt(Lcos(θ) / g)
Physics
1 answer:
Brrunno [24]3 years ago
7 0

Answer:

b

Explanation:

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Answer:

I believe it is C. Their Temps.

Explanation:

Hope my answer has helped you!

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Why centre of mass equal to centre of gravity
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Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

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3 years ago
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4 0
3 years ago
The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium
Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz

Given:

Energy = 378 \frac{kJ}{mol}

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × 10^{3} = 6.63 × 10^{-34} × v

v = 111.31 × 10^{37} Hertz

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz


7 0
3 years ago
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