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iren [92.7K]
3 years ago
15

Given the distance from the through of one wave to the next wave you can determine the HELP FAST!!!!!

Physics
1 answer:
makvit [3.9K]3 years ago
5 0

you can determine the wavelength

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When the voltage across a steady resistance is doubled, the current?
natima [27]

I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

The next question is what I'm not sure about:

Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).

What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

8 0
3 years ago
What happens if :<br> . The test charge is not tiny.
docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0
2 years ago
Where did we use rotational and irrotational flow​
WARRIOR [948]

Answer:

The term rotational and irrotational flow is associated withe the flow of particles in fluid.

The common example of irrrotational flow can be seen on the carriages of the Ferris wheel (giant wheel).

Explanation:

  • If the fluid is rotating along its axis with the streamline flow of its particles,then this type of flow is rotational flow.
  • Similarly if fluid particles do not rotate along its axis while flowing in a stream line flow then it is considered as the irrotational flow.
  • In majority, if the flow of fluid is viscid then it is rotational.
  • Fluid in a rotating cylinder is an example of rotating flow.
3 0
3 years ago
In a perfectly elastic collision between two perfectly rigid objects
ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

mu+MU = mv+MV

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

LearnwithBrainly

7 0
3 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
Read 2 more answers
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