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vladimir1956 [14]
3 years ago
10

a sample of a gaseous substance changes into liquid. This process is called (blank) and it takes place when heat is (blank) a ga

seous system
Physics
1 answer:
Delicious77 [7]3 years ago
8 0
Process is called condensation and it happens when heat is removed from gaseous system.
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Two planets are orbiting a star. Planet A is closer to the star than Planet B. Which description explains the expected motion of
77julia77 [94]
<span>Planet A orbits faster than Planet B, i believe is the answer. hope that helps

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50POINTS! Find the gravitational force between Earth and Venus - look up the data that you need for this problem.
SOVA2 [1]

The gravitational pull between Earth and venus is 831m

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3 years ago
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A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move
DochEvi [55]

Answer:

W=-2.1405\times 10^9\,J

Explanation:

Given:

electric field, E=148\,V.m^{-1}

charge, Q=-13\,\mu C=-13\times 10^{-6}\,C

initial position coordinates, p1 =(-18,-131)

final position coordinates,   p2 =(107,76)

We find the distance through which the charge has been moved:

d=\sqrt{(x1-x2)^2+(y1-y2)^2}

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

d=\sqrt{(107-(-81))^2+(76-(-131))^2}

d= 279.63\,m

Now we need the angle through which displacement is made with respect to the direction of electric field.

tan\,\theta= \frac{y2-y1}{x2-x1}

\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]

\theta= 47.75^{\circ}

Now from the relation between the change in potential difference:

\Delta V= E.d.cos\,\theta

\Delta V= 148\times 279.63\times cos\,47.75^{\circ}

\Delta V= 27826.06 V

∵The change in voltage is defined as the work done per unit charge.

∴\Delta V=\frac{W}{Q}

W=\frac{\Delta V}{Q}

Putting the respective values

W=\frac{27826.06 }{-13\times 10^{-6}}

W=-2.1405\times 10^9\,J

3 0
3 years ago
Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro
luda_lava [24]

Answer:

c. 4.582\times10^{21} kg

Explanation:

r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

v_{i} = Initial speed of rock = 136 ms⁻¹

v_{f} = Final speed of rock = 392 ms⁻¹

m = mass of the rock

M = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

8 0
4 years ago
Can someone please help me with this question thank you!
katovenus [111]
<h2>at the highest point that is B</h2><h2>because PE is directly proportional to height </h2><h2>hope it helped</h2>

7 0
3 years ago
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