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4 years ago

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A 0,2 kg ball is whirled at constant angular velocity in a vertical circle at the end of a 1.25m long string. If the maximum ten

sion the string can stand is 6N, what is the maximum angular velocity a ball can have in radians per second?

1 answer:

Dmitrij [34]4 years ago

8 0

Answer:

24 rad/s

Explanation:

.................

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The force of attraction between two objects is proportional to the mass of the objects and inversely proportional to the square of teh distance between them. In this casem if the distance is decreased to 1/10 of the original, then teh force of attraction increases by 100 times (10^2).

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What do scientist use the H-R Diagram for and how do they use it?

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3 years ago

Calculate the work function that requires a 475 nm photon to eject an electron of 1.25 eV.

lapo4ka [179]

The work function has the formula
E = h v

where
E is the energy
h is the constant
v is the frequency

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v = c / λ
where
c is the speed of light
<span>λ is the wavelength
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So.
E = h c / <span>λ
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3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates at 1.5m/s2? A. 51.5 N B. 4

const2013 [10]

Use the Second Law of Newton, which states that Net Force = mass times acceleration

F = m * a

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Answer: option C

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3 years ago

A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of

Aleksandr [31]

Answer:

Part 1)

$L_1 = 185.2 kg m^2/s^2$

Part 2)

$L_2 = 663.07 kg m^2/s^2$

Part 3)

$L = 663.07 kg m^2/s^2$

Part 4)

$\omega = 1.83 rad/s$

Part 5)

$F_c = 453.6 N$

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

$L_1 = I \omega$

here we know that

$I = 194 kg m^2$

$\omega = 1.47 rad/s^2$

now we have

$L_1 = 194 \times 1.47$

$L_1 = 185.2 kg m^2/s^2$

Part b)

Angular momentum of the person is given as

$L = mvR$

so we have

$m = 68 kg$

$v = 4.9 m/s$

R = 1.99 m

so we have

$L_2 = (68)(4.9)(1.99)$

$L_2 = 663.07 kg m^2/s^2$

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

$L = 663.07 kg m^2/s^2$

Part 4)

By angular momentum conservation of the system we will have

$L_1 + L_2 = (I_1 + I_2)\omega$

$185.2 + 663.07 = (194 + 68(1.99^2))\omega$

$848.27 = 463.28 \omega$

$\omega = 1.83 rad/s$

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

$F_c = m\omega^2 R$

$F_c = 68(1.83^2)(1.99)$

$F_c = 453.6 N$

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3 years ago