Answer:
True
the reason why I chose tire is because when you put a straw in a cup of water the water refract or bend
We have that the tension in rope 2 is mathematically given as
the tension in rope 2 is
From the question we are told
A crate hangs from a rope that is attached to a metal ring.
The metal ring is suspended by a <em>second </em>rope that is attached <u>overhead </u>at two points, as shown.
What is the angle if the tension in rope 1 is 0.640 times the tension in rope 2?
<h3> Angle of the tension</h3>
Generally the equation for the <em>Tension</em> is mathematically given as
Therefore
the tension in rope 2 is
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Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
Now At xp we have:
Which is a second order equation, using the quadratic formula to solve for xp would give us:
or
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that
BOTH are correct. This is simply explained by considring the following.
Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.
1) You can find the velocity with which the jumper leaves the ground by using the formula for the final velocity of an upward vertical motion.
Vf^2 = Vo^2 - 2gd
The data known are: Vf = 0, g = 9.81 m/s^2, and d = 60 cm = 0.6 m
=> Vo^2 = 2gd = 2*9.81m/s^2 * 0.6 m = .11.772 m^2/s^2 =>
Vo = 3.4 m/s
2) You can obtain the acceleration of the jumper during the jump by using the equivalent formula, Vf^2 = Vo^2 + 2ad
Where, Vf^2 is the velocity with which he leaves the ground (3.4 m/s), Vo is zero because he jumps from the rest, d is the distance that his body rises from knees,which is 50 cm = 0.5 m
Then, a = Vf^2 / 2d = [3.4m/s]^2 / (2*0.5m) = 11.6 m/s^2
And now, you can find the force exerted by the jumper as:
F = m*a, and if you know his weight, W = m*g => m = W/g
Replace m in F = m*a => F = W*a/g = W*11.6m/s^2 / 9.81 m/s^2 = 0.10 W
F = 0.10 W
ABSTRACT
Once the universe was created by the Big Bang, the only abundant elements present were hydrogen (H) and helium (He). These elements were not evenly distributed throughout space, and under the influence of gravity they began to "clump" to form more concentrated volumes. Evidence of this uneven distribution can be found in the anisotropies detected in the Cosmic Background Radiation (CMB) by the COBE satellite in the early 90's. These clumps would eventually form galaxies and stars, and through the internal processes by which a star "shines" higher mass elements were formed inside the stars. Upon the death of a star (in a nova or a supernova) these high mass elements, along with even more massive nuclei created during the nova or supernova, were thrown out into space to eventually become incorporated into another star or celestial body.
The conditions inside a star that allow the formation of the higher mass elements can be related to a pushing match between gravity and the energy released by the star. Gravity creates a force that would cause a star to shrink and collapse, but the energy released by nuclear reactions within the star flows outward, and produces thermal pressure that opposes gravity. When these two forces are balanced, the star maintains a particular size. But when there is some type of imbalance, the star (or some part of it) will expand or contract in response to the stronger of the two forces.
When the universe was first created, essentially all matter was in the form of two elements- hydrogen and helium. Their relative abundance (by weight) was 75% hydrogen and 25% helium. (This means that for every He nucleus there were 12 H nuclei/protons) They were not evenly distributed throughout space. This is critical, because this uneven distribution allowed gravity to act in the areas of higher concentration to initiate the "clumping" of matter. If everything were evenly spread out, nothing would have happened, for each atom would have been attracted evenly from all directions, and would have remained where it was relative to neighboring atoms. As a result of slight discrepancies in the distribution of matter, gravity was able to initiate the collapse of huge volumes of H and He into more concentrated areas of gas. These areas eventually would evolve to form galaxies. Within these areas, there was a second level of more concentrated clumping of H and He that would form stars, where the higher mass elements would be created.
In these more concentrated areas, as the clouds of H and He (called nebulas) collapsed, the atoms were speeding up as they were pulled toward the center by gravity. This caused two things to happen. First, the increase in the velocity of the atoms resulted in an increase in the temperature of the material. At some point, the temperature became high enough so that the material began to glow. Second, the atoms were becoming packed more tightly, increasing the density, and the frequency of collisions between atoms. As this happened, the mass of H and He became more spherical. At this stage the mass of H and He is called a protostar.
or copy and paist http://aether.lbl.gov/www/tour/elements/stellar/stellar_a.html