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3 years ago

The diagram shows a ball resting at the top of a hill.

Physics

2 answers:

andrey2020 [161]3 years ago

8 0

Answer:

The potential energy in the system is greatest at X.

The first option

WARRIOR [948]3 years ago

7 0

Answer:

a) The potential energy in the system is greatest at X.

Explanation:

Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

$U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}$

Conclusions are showed as follows:

a) The potential energy in the system is greatest at X.

b) The kinetic energy is the lowest at X and Z.

c) Total energy remains constant as the ball moves from X to Y.

Hence, the correct answer is A.

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Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

Semenov [28]

For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

6 0

4 years ago

Please help!!

Harlamova29_29 [7]

Answer:

This could represent something like sliding a small rock across an icy lake.

Explanation:

A 20N force of gravity (weight), and 20N normal force exerted back onto the object imply it is on the ground and has no vertical motion. There is a net force of 0N

An 80N force to the left and a 5N force to the right create a net force of 75N to the left. This means that there is a force acting on the object that makes it accelerate to the left. 80N represents a push or pull force and 5N represents a relatively small frictional force due to the object being slid on a surface such as steel or in this case ice.

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4 years ago

How is water vapour different from a mixture of hydrogen and oxygen atoms?

Pavel [41]

Answer: in water vapor hydrogen atoms are bonded to oxygen atoms. In a mixture of hydrogen and oxygen hydrogen atoms are bonded to hydrogen atoms and oxygen atoms are bonded to oxygen atoms.

Explanation:

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3 years ago

Instead, suppose that it was ﬁred upward at 60◦ with respect to a horizontal line. Then its horizontal component of velocity is

larisa [96]

Answer:

50 m/s

Explanation:

Angle = 60 degree

Horizontal component of velocity = 50 m/s

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When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

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3 years ago

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ELEN [110]

Helium is less dense than normal air

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3 years ago