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bagirrra123 [75]

3 years ago

15

P7.36 A ship is 125 m long and has a wetted area of 3500 m2. Its propellers can deliver a maximum power of 1.1 MW to seawater at

20C. If all drag is due to friction, estimate the maximum ship speed, in kn.

1 answer:

Mariana [72]3 years ago

4 0

The solution is in the attachment

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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

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What type of clouds are associated with low pressure?

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Somebody please help me with this question..!

Rudik [331]

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8.2) Difference in the shape of lines is due to different density of air & water. So, that physical property and your answer would be option A

So, in Short Answers of your questions are:
8.1) - Option D
8.2) - Option A

Hope this helps!

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3 years ago

Verify that the SI unit of impulse is the same as the SI unit of momentum.

lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

$P = mv$

Where:

P = momentum

m = mass

v = velocity

The SI unit:

$mass = kg\\ velocity = \dfrac{m}{s}$

So the unit of momentum would be:

$kg.\dfrac{m}{s}$

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or $kg.\dfrac{m}{s^{2} }$

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = $kg.\dfrac{m}{s^{2} }$ x $s$

or

$\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}$

You can then cancel out one s each from the numerator and denominator and you'll be left with:

$kg.\dfrac{m}{s}$

So then:

Momentum: Impulse

$kg.\dfrac{m}{s}$ $kg.\dfrac{m}{s}$

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3 years ago