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2 years ago

By quadrupling the mass of the body hanging on the thread what will become the period of oscillation?

Physics

1 answer:

nikklg [1K]2 years ago

6 0

Well, the tension in the thread will probably quadruple, but the hanging body will continue to just hang there.

The question gives us no evidence that it is doing any oscillating, and there's no reason for it to start just because it suddenly got heavier.

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What is refraction?

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B- light bends as it passes through an object ( a would be reflect)

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2 years ago

A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

$N_A=6.022\cdot 10^{23} mol^{-1}$ is the Avogadro number

Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

The energy emitted by each atom (the energy of one photon) is

$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda=508 nm=5.08\cdot 10^{-7}nm$ is the wavelength

Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

And so, the total energy emitted by the sample is

$E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J$

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2 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

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Answer:

Explanation:

To solve this, we start by using one of the equations of motion. The very first one, in fact

V = U + at.

V = 0 + 0.8 * 3.4 = 2.72 m/s.

V = 0 + 0.8 * 4.3 = 3.44 m/s.

d = ½ * 0.8 * 4.3² + 3.44 * 12.9

d = 7.396 + 44.376

d = 51.77 m.

d = 62 - 51.77 = 10.23 m. = Distance

traveled during deceleration.

a = (V² - Vo²) / 2d.

a = (0² - 3.44²) / 20.46

a = -11.8336 / 20.46 = -0.58 m/s²

t = (V - Vo)/a =(0 - 3.44) / -0.58

t = -3.44/-.58 = 5.93 s

= Stop time.

T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total

time the hare was moving.

d = Vo * t + ½ * a * t² = 62 m.

0 + 0.5 * (23.13)² * a = 61

267.5a = 61

a = 61/267.5

a = 0.23 m/s²

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2 years ago

A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

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I had the same question

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2 years ago

Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo

Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

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#SPJ4

8 0

1 year ago

By quadrupling the mass of the body hanging on the thread what will become the period of oscillation?​

By quadrupling the mass of the body hanging on the thread what will become the period of oscillation?