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3 years ago

Hanging from a horizontal beam are nine simple pendulums of the following lengths:

Physics

1 answer:

LenaWriter [7]3 years ago

8 0

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

$L =\frac{g}{w^{2} }$

where g=9.8m/s

ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

$L= \frac{9.8}{2^{2} }$

L = 2.45m

when ω= 4rad/sec

$L=\frac{9.8}{4^{2} }$

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

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Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0

AURORKA [14]

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

5 0

3 years ago

A rope vibrates every 0.5 s. What is the frequency of the waves? 5 Hz 2 Hz 0.5 Hz 1 Hz

yulyashka [42]

The answer is 2Hz

Using the formula f= 1/T we can plug in .5 for T and solve for frequency.

4 0

3 years ago

Which of the following cannot be determined from the spectrum of a star?

Aneli [31]

I think the correct answer from the choices listed above would be the last option. It is the chemicals in the core of the star that cannot be determined from the spectrum of a star. Spectrum shows the different classification of the stars depending on their spectral characteristics. It usually involves the light, the wavelength and the distance.

4 0

3 years ago

Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of

MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of $Cl_2$ = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of $Cl_2$ = 71 g/mole

Molar mass of $PCl_5$ = 208.24 g/mole

First we have to calculate the moles of $P$ and $Cl_2$ .

$\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles$

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2P+5Cl_2\rightarrow 2PCl_5$

From the balanced reaction we conclude that

As, 5 moles of $Cl_2$ react with 2 moles of $P$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $P$

That means, in the given balanced reaction, $Cl_2$ is a limiting reagent and it limits the formation of products and $P$ is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of $PCl_5$ .

As, 5 moles of $Cl_2$ react with 2 moles of $PCl_5$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $PCl_5$

Now we have to calculate the mass of $PCl_5$ .

$\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5$