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3 years ago

7

Which type of wave interaction is shown in the diagram?

1 answer:

masya89 [10]3 years ago

5 0

Answer: Constructive interference

Explanation: Just took the test

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Which of these attributes is generally attributed to Kepler

kotegsom [21]

Kepler's first "law":
The orbits of the planets are ellipses, with the sun at one focus.
This would include Mars.

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4 years ago

Planet that has an orbit that intersects the orbit of pluto

Scilla [17]

<h2>Answer: Neptune </h2>

The dwarf planet Pluton, <u>has the most eccentric orbit</u> (more elliptical and elongated) of all the planets, and as a consequence its orbit is "intersected" by the orbit of Neptune.

However, <u>despite this intersection, there is no collision risk between these two bodies</u>, since the orbit of Pluto is located in an orbital plane different from that of the other planets and therefore different from that of Neptune, its nearest neighbor. In addition, the orbit of Pluton is inclined $17\°$ on the the ecliptic plane (plane where the other planets move around the Sun).

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3 years ago

What can reduce a machines efficiency

algol [13]

Answer:

its power.

Explanation:

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3 years ago

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Un cuerpo recibe una fuerza de 100N y produce una aceleracion de 2 m/s² determine la masa del cuerpo

kykrilka [37]

Answer:

The mass of the body is 50 kg.

Explanation:

We have,

Force acting on a body is 100 N

Acceleration of the body is 2 m/s²

It is required to find the mass of the body. The force acting on a body is given by :

F = ma

$m=\dfrac{F}{a}\\\\m=\dfrac{100}{2}\\\\m=50\ kg$

So, the mass of the body is 50 kg.

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3 years ago

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

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