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inna [77]
3 years ago
11

How often do spring and neap tides occur

Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

Without thought to the season, Spring tides happen [two times] every [lunar] month the whole year. When the Sun and the Moon are quadratic, Neap tides also happen two times during a month.

Here is a picture to help show you:

(With the Spring tide, the moon is toward the "left" or "right" of the Earth. With the Neap tide, the moon is "above" or "below/under" the Earth. It's kind of hard to explain).

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Radio waves just like light waves can be reflected refracted and diffracted and polarized.
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<span>Radio waves just like light waves can be reflected refracted and diffracted and polarized.  The answer is True. </span>These characteristics are the common phenomena for electromagnetic (EM)  waves, and Radio Waves are electromagnetic Waves so much so that they obey reflection, refraction, and diffraction. 
5 0
3 years ago
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What will be the change in momentum caused by a net force of 120N acting on an object for 2 seconds?
Vinvika [58]

The change in  momentum is 240 kgm/s

<u>Explanation:</u>

Given:

Force, F = 120N

Time, dt = 2 sec

Change in momentum, dP = ?

We know,

F = \frac{dP}{dt}

On substituting the value we get:

120 = \frac{dP}{2}\\ \\dP = 240 kgm/s

Therefore, change in  momentum is 240 kgm/s

8 0
3 years ago
1. A skier takes off from the top of the ski run and
vampirchik [111]

Answer:

a = 6 [m/s^2]

Explanation:

In order to calculate the acceleration of the skier, the following expression of kinematics must be used:

a = (v)/t

where:

v = velocity = 24 [m/s]

t = time = 4 [s]

a = 24/4 = 6 [m/s^2]

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3 years ago
How many grams of sugar are in 0.5mol C6H12O6?
Anastasy [175]

Answer:

90.07794 grams

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5 0
3 years ago
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A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
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