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vovikov84 [41]
4 years ago
13

Water is observed spilling to the floor in a spaceship. Describe the motion and location of the spaceship. Use the behavior of t

he water to justify your answer.
Physics
2 answers:
Mashutka [201]4 years ago
8 0

The spaceship is either not moving and is in its launch pad here on earth and the water is behaving as it would anywhere else on our planet

or

the spaceship is in outer space somewhere and it has fired up its thrusters and it has created it's own gravitational Force because it is accelerating

or

the spaceship is trying to land on another planet and is slowly coming to the surface of the planet

or

the spaceship is taking off from a planet and the gravitation force of the planet is still in effect.

makkiz [27]4 years ago
4 0

Answer:

Based on the principle of equivalence, the motion and location of the spaceship are unknown.

The ship may be accelerating in space, where the floor rises up to the water.

The spaceship may be stationary on Earth, where gravity causes the water to fall to the floor.

Explanation:

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Two energy transfers take place when a book hits the ground, Which type of energy transfers are those
bazaltina [42]

Answer:

I think that when a book hits the ground its potential energy converts into kinetic energy and then kinetic energy is transformed into sound and heat energy.

Explanation:

6 0
3 years ago
A non uniform rod has mass
Doss [256]

Answer:

r_{cm} = L/3

Explanation:

Mass: M, Length: L.

\sigma (x) = b(L-x)

The formula that gives center of mass is

\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}

In the case of a non-uniform mass density, this formula converts to

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.

Here x's are cancelled. Otherwise, the denominator would be zero.

r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}

8 0
3 years ago
A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass is 57.0
musickatia [10]

Answer:

568.16J

Explanation:

V = 1.4 m/s

h = 2.15m

H = 12.4m

M = 57.0kg

Fr = 41.0N

Vb = 6.80m/s

g = 9.8m/s²

K.E at the top + P.E at the top + Work done = K.E at the bottom + frictional force.

½mv² + mgh + W = ½mVb + Fr.H

(½* 57 * 1.4²) + (57 * 9.8 * 2.15) + W = (½ * 57 * 6.8²) + (41 * 12.4)

55.86 + 1202.22 + w = 1317.84 + 508.4

1258.08 + w = 1828.24

W = 1826.24 - 1258.08

W = 568.16J

5 0
3 years ago
Imagine that a loudspeaker is producing a quiet note with a low pitch. How will its vibrations change:
borishaifa [10]

Answer:

See below

Explanation:

Higher pitch mens higher frequecy of vibrations

 louder means more amplitude of vibrations (stronger vibrations)

3 0
2 years ago
A plane cruising at 233 m/s accelerates at 17 m/s 2 for 4.8 s. What is its final velocity? Answer in units of m/s. 013 (part 2 o
Volgvan

Answer:

Final velocity will be 314.6 m/sec

Distance traveled = 1314.24 m

Explanation:

We have given initial velocity u = 233 m/sec

Acceleration a=17m/sec^2

Time t = 4.8 sec

From first equation of motion v=u+at, here v is final velocity, u is initial velocity and t is time

So v=233+17\times 4.8=314.6m/sec

Now we have to find distance traveled

From second equation of motion

S=ut+\frac{1}{2}at^2=233\times 4.8+\frac{1}{2}\times 17\times 4.8^2=1314.24m

So distance traveled in given time will be 1314.24 m

4 0
3 years ago
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