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solong [7]
3 years ago
10

Path:p

Physics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

.1 A1 = a1      A1 is amount at .1 %

.01 A2 = a2    A2 is the amount at .01%

a1 + a2 = 454     total of a1 and a2

A = 454 / .03  =  15133.33    tot amount needed

454 = .1 A1 + .01 A2    

A2 = 15133.33 - A1    

454 = .1 A1 + .01 (15133.33 - A1)

454 = .09 A1 + 151.33

A1 = 302.67 / .09 = 3363

A2 = 15133.3 - 3363 = 11770.3

Check:

.1 * A1 = 336.3

.01 * A2 = 117.7

total = 336.3 + 117.7 = 454

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MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1  = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

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After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

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(c)

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(d)

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Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

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5 0
2 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

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What is the best unit to use when measuring the mass of a mineral sample?
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A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths
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Answer:

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d = (1 - 1/2) (531 / (2)(1.34)

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3 0
3 years ago
(a) If a long rope is hung from a ceiling and waves are sent. up the rope from its lower end, why does the speed of the waves ch
vazorg [7]

My response to question (a) and (b) is that all of the element of the rope need to  aid or support the weight of the rope and as such, the tension will tend to increase along with height.

Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ   increases with height.

<h3>How does tension affect the speed of a wave in a rope?</h3>

The Increase of the tension placed on a string is one that tends to increases the speed of a wave, which in turn also increases the frequency of any given length.

Therefore,  My response to question (a) and (b) is that all of the element of the rope need to  aid or support the weight of the rope and as such, the tension will tend to increase along with height.  Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ   increases with height.

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See full question below

(a) If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, why does the speed of the waves change as they ascend? (b) Does the speed of the ascending waves increase or decrease? Explain.

4 0
2 years ago
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