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deff fn [24]
3 years ago
6

What is the difference between a neap and a spring tide in terms of size?

Physics
1 answer:
Mamont248 [21]3 years ago
7 0
Spring tides occur when the moon is either new or full, and the sun, the moon, and the Earth are aligned. ... neap tide- A tide in which the difference between high and low tide is the least. Neap tides occur twice a month when the sun and moon are at right angles to the Earth.
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17. Saan daw nakasakay ang mga Austronesyan nang dumating sa bansa? A. Bangka B. balangay C galyon D. barko
erastova [34]

Answer:

D.barko po

Explanation:

I trying my best

7 0
2 years ago
Consider dropping a ball from rest. This ball moves from astate of high gravitational potential energy to one of lowgravitationa
Leni [432]

Answer:

From the negative to the positive cable.

Explanation:

The electrons have negative charge, which means that the negative terminal of the battery will suply the electrons, thus they are present in excess on the negative cable and will jump from it to the positive cable. This current direction is called real current.

3 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
What is one way to increase the amplitude if a wave in a medium?
Gekata [30.6K]

Answer: c

Explanation:

3 0
3 years ago
A student throws a rock upwards. The rock reaches a maximum height 2.4 seconds after it was released.
Allisa [31]

Answer:

23.52 m/s

Explanation:

The following data were obtained from the question:

Time taken (t) to reach the maximum height = 2.4 s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)

was thrown as follow:

v = u – gt (since the rock is going against gravity)

0 = u – (9.8 × 2.4)

0 = u – 23.52

Collect like terms

0 + 23.52 = u

u = 23.52 m/s

Therefore, the rock was thrown at a velocity of 23.52 m/s.

7 0
3 years ago
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