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3 years ago

6

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

1 answer:

Anastasy [175]3 years ago

6 0

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

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How does the physical structure of a plant have<br> Large leaves help it survive?

elena-14-01-66 [18.8K]

The large leaves help it survive as they serve as the<u> organ for photosynthesis.</u>

Explanation:

Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy.
During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds

Leaves provide food and air to help a plant stay healthy and grow. Through photosynthesis, leaves turn light energy into food.
Through pores, or stomata, leaves breathe in carbon dioxide and breathe out oxygen. Leaves also release excess water.
Most leaves are broad and so have a large surface area allowing them to absorb more light
A thin shape means a short distance for carbon dioxide to diffuse in and oxygen to diffuse out easily.
The exchange of oxygen and carbon dioxide in the leaf occurs through pores called stomata.
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3 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

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3 years ago

Which theory of language development and acquisition includes the concepts of reinforcement, stimuli control, deprivation, and a

Vlada [557]

A skinner idea of language .-.

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4 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{42}\textrm{K}$ = 41.962403 u

Mass of $_{20}^{42}\textrm{Ca}$ = 41.958618 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)\times c^2$

$E=(0.003785u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=3.526MeV$

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

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4 years ago

Which statement best describes energy and matter in a closed system? (2 po

Ivahew [28]

Answer:

Energy can flow into and out of the system but matter cannot.

Explanation:

In a closed system, energy can flow in and out of the system but matter cannot.

A closed system prevents double way flow of matter.
A closed system conserves matter.

For an isolated system, energy and matter cannot flow out of the system.

For open systems, energy and matter can flow out of the system.

Such systems are used for certain thermodynamics experiment.

4 0

3 years ago