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3 years ago

If tides occur twice a day in most places. How much time is there between a morning high tide and the next high tide?

Physics

1 answer:

algol133 years ago

8 0

About 12 hours is the time between a morning high tide and the next high tide

Explanation:

The Earth’s rotation happens between two tidal bulges

The “periodic rise and fall” of the surface water levels of the ocean is called tides. The gravitational action and interaction on the earth by the sun and the moon causes these tides. Different regions of the World experiences different patterns of tides like the diurnal, semi-diurnal etc.

When there is one high and one low tide occurring on a lunar day, then it is diurnal pattern. Semi-diurnal pattern occurs when there are two equal high and low tides on a single lunar day.

Since the Earth’s rotation happens between two tidal “bulges” on each lunar day, the coastal areas can experience two high and two low tides in every 24 hours plus 50 minutes.

Accordingly the time between two high tides would be 12 hours plus 25 minutes. Similarly, the time gap between a high to low tide would be 6 hours plus 12.5 minutes.

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A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

2 years ago

A certain metal wire has a cross sectional area of 1 mm2 and is 1 m long. when it is hung from the ceiling and a 10 kg mass is h

kvasek [131]

From the Hooke's law , the extension force of an elastic material is directly proportional to the extension.
That is, F = k e, where F is the force , k is the constant and e is the extension
F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
= 100 N/ 0.001
= 100000 N/m or 100 N/mm

5 0

3 years ago

A car travel with a constant velocity of 20.5m/s for 20seconds what distance does it cover in this time ?

prohojiy [21]

Answer:

410 m

Explanation:

Given:

v₀ = 20.5 m/s

a = 0 m/s²

t = 20 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²

Δx = 410 m

8 0

3 years ago

Which of the following is a correct equation for total energy?

Liula [17]

Answer:

kinetic energy + potential energy

6 0

3 years ago

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If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *

Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0

3 years ago

Read 2 more answers