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3 years ago

Describe how refraction concentrates wave energy so that some parts of a beach erode more.

1 answer:

3 0

Refraction is the bending of the waves which is result of the fact that different parts of the wave reach the water with different speeds because of the angle approaching the shore.

The wave refraction disperses the wave energy in quiet water areas and sand is deposited.

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A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.

victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀ $e^{-\lambda t}$

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ $e^{-\lambda_1 t}$

For B

N(B) = N(B)₀ $e^{-\lambda_2 t}$

N(A) = N(B) , for t = 2 h

N(A)₀ $e^{-\lambda_1 t}$ = N(B)₀ $e^{-\lambda_2 t}$

N(A)₀ $e^{-\lambda_1 t}$ = 5 x N₀(A) $e^{-\lambda_2 t}$

$e^{-\lambda_1 t}$ = 5 $e^{-\lambda_2 t}$

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

Putting t = 2 h , λ₁ = .9 h⁻¹

$e^{\lambda_2\times 2}$ = 5 $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0

3 years ago

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V batte

Ugo [173]

Answer:

1 μC extra charge will be flow here

Explanation:

Given data

battery V1 = 4.0 V

flows Q1 = 6.0 μC

replace battery V2 = 7.0 V

to find out

what happen if we replace battery

solution

we apply here principal of capacitor

that is Q directly proportional voltage

so we say Q2/Q1 = V2/ V1

put all value here

Q2/Q1 = V2/ V1

Q2/6 = 7/ 6

Q2 = 7

so we see here 7 μC will be flow

and Q = Q2 - Q1 = 7 - 6 = 1 μC

so we also say that 1 μC extra charge will be flow here

5 0

3 years ago

Evaporates can provide an ideal speed racing surface. True or False

musickatia [10]

the answer is true because evaporates can provide an ideal speed racing surface

6 0

3 years ago

Jalen combines two substances in science class. Once combined, only one of the two substances

dusya [7]

Answer: b. The combination is a mixture because the substances can be separated

Explanation: Based on the facts presented above, the combination of both both substances can be referred to as a mixture due to the following:

A mixture is obtained when two or more substances or materials are combined without a chemical reaction. This is observed when Jalen combined substance 1 and 2 with only one of the substances becoming visible after the combination.

The other reason is that, a mixture can be separated back into its original constituent, this is evident when the combination was filtered with only substance 2 going through the filter and substance 1 remaining in the filter

4 0

3 years ago