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katen-ka-za [31]
3 years ago
7

What is an atomic nucleus?

Physics
1 answer:
AlexFokin [52]3 years ago
3 0

It is basically a very tiny region centering around protons and neutrons in the center of a atom

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Which statement is the most appropriate conclusion for David to write?
ycow [4]
I need more to answer this
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3 years ago
A ball is dropped from rest at height of 20m. If it loses 25% of its kinetic energy when it strikes the ground. What is the heig
victus00 [196]

Answer:

15 meters

Explanation:

The inicial energy of the ball is just potencial energy, and its value is:

E = m * g * h = m * g * 20,

where m is the ball mass, and g is the value of gravity.

In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:

E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g

This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.

So, as the mass and the gravity are constants, we can calculate the height the ball will reach:

E' = m*g*h = 15*m*g -> h = 15 meters

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3 years ago
Earth, has a bigger mass than the Moon .
raketka [301]
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3 years ago
We can determine the velocity of a wave when given the _______ and the wavelength.
ehidna [41]

Answer:

velocity = frequency * wavelength

Given the wavelength one must also have the frequency.

7 0
3 years ago
A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.
IrinaVladis [17]

Answer:

Approximately 722\; \rm m\cdot s^{-1}.

Explanation:

The average speed of a vehicle is calculated as:

\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}.

In this question, the total distance is 2 \times 560\; \rm km = 1120\; \rm km.

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m.

1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m.

Time required for the first part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s.

Time required for the second part of this trip:

\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s.

The time required for the entire trip would be approximately 930 + 620 = 1550\; \rm s.

Calculate the average speed of this plane:

\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}.

6 0
3 years ago
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