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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

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

</span>

6 0

3 years ago

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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

4 0

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