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kolbaska11 [484]
2 years ago
12

Ocean waves are hitting a beach at a rate of 3.2 Hz. What is the period of the waves? (2 decimal places)

Physics
1 answer:
insens350 [35]2 years ago
7 0

Answer:

0.31s

Explanation:

Given parameters:

Frequency of the ocean waves  = 3.2Hz

Unknown:

Period of the waves  = ?

Solution:

The period of a wave is the inverse of the frequency of a wave;

       Period  = \frac{1}{Frequency}

Now insert the parameters and solve;

       Period  = \frac{1}{3.2}   = 0.31s

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A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at
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Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to 1 × 10^{-12} meter

1 micrometer is equivalent to 1 × 10^{-6} meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by 174 × 10^{-12} meter we get that the number of sheets will be as follows

\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719

Therefore, they are approximately 719 sheets

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3 years ago
You are asked to determine the density of a liquid. What lab equipment is most helpful to find density of the liquid?
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D. The graduated cylinder is used to find the volume, and triple beam balance is used to find the mass.
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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu
xeze [42]

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

8 0
2 years ago
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