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3 years ago

What is the difference between a low tide and a high tide

Physics

2 answers:

Savatey [412]3 years ago

6 0

A high tide means when the water has risen and is higher up(closer to high up land). Low tide is when it’s receded

Oxana [17]3 years ago

5 0

Answer:

High water level during a tide is called High tide.

Low water level during a tide is called Low tide.

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A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.

grandymaker [24]

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically, I = $\frac{P}{A}$

As it is given that power is 28.0 W and area is $7 \times 10^{-5} m^{2}$ .

Therefore, sound intensity will be calculated as follows.

I = $\frac{P}{A}$

= $\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}$

= $0.318 \times 10^{5} W/m^{2}$

or, = $3.18 \times 10^{4} W/m^{2}$

Thus, we can conclude that sound intensity at the position of the microphone is $3.18 \times 10^{4} W/m^{2}$ .

7 0

3 years ago

A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?

zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0

3 years ago

Can someone help quickly please and thank you

Sergio039 [100]

Answer:

I think D??

Explanation:

8 0

2 years ago

Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?

iren2701 [21]

Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.

8 0

3 years ago

200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.

balu736 [363]

Answer:

a. -0.01324 kJ/K, b. = 0.03233 kJ/K , c. = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)

3.54 = 0.1 . X x 203.29

x = 0.1741, which is the dryness fraction of fluid at the final state.

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= 20.329 kJ, which is much greater than 3.45 kJ, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.

6 0

3 years ago