All categories

3 years ago

Explain the difference between a high tide and a low tide.

Physics

1 answer:

SIZIF [17.4K]3 years ago

4 0

Answer: Tidal range

Explanation:

Tides are considering the rise and fall of sea levels and there are two types of it which are called high tide and low tide. The difference between high tide and low tide is called the tidal range.

The tidal range is not constant and it is considering height difference. It can change and it is depending on the locations of the Sun or the Moon.

High tide is the highest level of the place where the water rises because when the water rises to its highest level, then the water is reaching its high tide.

When it comes to low tide, then it is the opposite of high tide. Water is reaching its lowest level.

You might be interested in

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

3 0

3 years ago

What s the rate of acceleration of gravity?

Ivenika [448]

The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.

4 0

3 years ago

Read 2 more answers

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

finlep [7]

In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".

5 0

3 years ago

This for my previous question.. im so sorry

Marta_Voda [28]

Answer:

the answer for the question is C

3 0

3 years ago