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4 years ago

15

The terrestrial planets are made almost entirely of elements heavier than hydrogen and helium. According to modern science, wher

e did the elements heavier than hydrogen and helium come from?

2 answers:

Viktor [21]4 years ago

6 0

Answer:

Pls refer to this website:

brainly.com/question/13449023

drek231 [11]4 years ago

4 0

Hydrogen makes stars and stars spend most of their lifetime making helium.

The heavier elements are made when an old-age, high mass star explodes as a nova or supernova and then dies.

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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji

WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is $-14.75^{o}$

Explanation:

Force by Jack in vector form

$\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)$

Force by Jill in Vector form is given by

$\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

Force by Jane is

$\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}$

Net force is:

$\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}$

Hence

$\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

The net force will be given by

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}$

The direction of net force is:

$\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}$

The angle with East direction is $-14.75^{o}$

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is $-14.75^{o}$ and it is $14.75^{o}$ from the south.

5 0

3 years ago

A boy drops a coin down a well that is 225 m deep. How long does it take the coin to hit the bottom of the well? please help

Nady [450]

<span>It takes 6.78 seconds for the coin to hit the bottom of the well. We can use the equation h = 0.5gt^2, where h is the height of the coin, g is the gravitational constant of 9.8m/s^2, and t is the time is takes for the coin to hit the bottom of the well. Solve for t to obtain 6.87 seconds.</span>

8 0

3 years ago

.Roshan applied a force of 144 N to the side of a cube-shaped wooden block to slide it away from himself . Assuming that he appl

ExtremeBDS [4]

side of cube=4cm

Area=side^2=4^2=16cm^2=0.0016m^2

Force=144N

$\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}$

$\\ \sf\longmapsto Pressure=\dfrac{144}{0.0016}$

$\\ \sf\longmapsto Pressure=90000Pa$

3 0

3 years ago

13. If you shorten the length of string by half that holds an object in rotation at the same tangential

Dmitrij [34]

13. doubles

The tension in the string corresponds to the centripetal force that holds the object in rotation, so:

$T=F=m\frac{v^2}{r}$

where m is the mass of the object, v is the tangential speed, and r is the distance of the object from the centre of rotation (therefore it corresponds to the length of the string). The problem tells us that the tangential speed remains the same (v), while the length of the string is halved, so r'=r/2. Therefore, the new tension in the string will be

$T'=m\frac{v^2}{r'}=m\frac{v^2}{r/2}=2m\frac{v^2}{r}=2T$

so, the Tension doubles.

14. Variations of centripetal forces

Both revolution and rotation refer to the rotational motion of an object, therefore they both involve the presence of a centripetal force, which keeps the object in circular motion. The only difference between the two is:

- Revolution is the circular motion of an object around a point external to the object (for instance, the motion of the Earth around the Sun)

- Rotation is the circular motion of an object around its centre, so around a point internal to the object (for instance, the rotation of the Earth around its axis)

15. Rotational speed

For a uniform object in circular motion, all the points of the object have same rotational speed. In fact, the rotational speed is defined as

$\omega=\frac{\Delta \theta}{\Delta t}$

where $\Delta \theta$ is the angular displacement covered in a time interval of $\Delta t$ . Since all the points of the wheel are coeherent (they move together), they all cover the same angular displacement in the same time, so they all have same rotational speed.

16. away from the center of the path.

The tension in the string is responsible for keeping the tin can in circular motion. Therefore, the tension in the string represents the centripetal force, and so it is directed towards the centre of the path. According to Newton's third law, the tin can exerts a force on the string which is equal in magnitude (so, same magnitude of the tension), but opposite in direction: therefore, away from the centre of the path.

17. weight of the bob.

There are two forces acting on the bob in the vertical direction: the weight of the bob (downward) and the vertical component of the string tension (upward). Since there is no acceleration along the vertical direction, the net force must be zero, so these two forces must be equal: it means that the vertical component of the string tension is equal to the weight of the bob. Along the horizontal direction, instead, the horizontal component of the string tension corresponds to the centripetal force that keeps the bob in circular motion.

18. horizontal component of string tension.

Along the horizontal direction, there is only one force acting on the bob: the horizontal component of the string tension. Since the bob is moving of circular motion along the horizontal direction, this means that this force (the horizontal component of the string tension) must correspond to the centripetal force that keeps the pendulum in circular motion.

19. inward, toward the center of swing.

The force that the can exerts on the bug is the force that keeps the bug in circular motion (since it prevents the bug from moving away). Therefore, it must corresponds to the centripetal force.

20. speed of the car. AND radius of curvature.

The normal force exerted on a car executing a turn on a banked track is given by the expression:

$N=\frac{mg}{cos \theta - \mu sin \theta}$

where m is the mass of the car, g is the gravitational acceleration, $\theta$ is the angle of the bank, and $\mu$ is the coefficient of friction.

From the formula, we see that the normal force depends on $\theta$ (the angle of the bank) and $\mu$ (the coefficient of friction), while it does not depend on the speed of the car or on the radius of curvature. Therefore, these two are the correct answers.

3 0

3 years ago

According to our present theory of solar system formation, why were solid planetesimals able to grow larger in the outer solar s

nirvana33 [79]

Answer:

According to our present theory of solar system formation, why were solid planetesimals able to grow larger in the outer solar system than in the inner solar system? Because only metal and rock could condense in the inner solar system, while ice also condensed in the outer solar system.

Explanation:

5 0

2 years ago