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ollegr [7]
3 years ago
13

A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What i

s its speed there, assuming energy is conserved?
105 m/s
Its speed will depend on how it got to the new place.
Less than 105 m/s
More than 105 m/s
Physics
1 answer:
Alex_Xolod [135]3 years ago
4 0

Answer:

More than 105 m/s

Explanation:

Electric potential is defined as the work done in moving a charge from one point to another in an electric field.

Since the charge moved from a region of higher potential (10V) to a region of lower potential (5V) the speed is bound to increase because lesser work will be required to keep it moving; this implies that the kinetic energy will increase since the same amount of energy that kept it moving in the 10V region is still supplied to it ( energy is conserved).

Another simple way to help understand this is that whenever the potential energy of a particle decreases its kinetic energy increases; which implies that its velocity must have increased also, as long as the total energy of the system is conserved.

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A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?
givi [52]
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

             Frequency  =  (speed)  /  (wavelength)

                                 =  (6 m/s)  /  (12 m)

                                 =      0.5 / sec

                                 =      0.5 Hz .  

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
5 0
2 years ago
Read 2 more answers
Which planet has a storm called the Great Dark Spot swirling in its atmosphere?
Viktor [21]
Im thinking D. Neptune..
3 0
3 years ago
Read 2 more answers
Putting a beaker of water in a microwave and turning it on for a minute or two will increase the temperature of the water. Descr
Jobisdone [24]

Answer:

The equipment to use is: a beaker, a fixed amount of water, a thermometer.

The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made

Explanation:

The design of an experiment is to place the beaker in the microwave, with a good amount of water (approximately ⅔ of its capacity) and turn it on for small periods of time, generally the minimum is 30 s, quickly open the microwave, place a thermometer or better yet an infrared thermometer to measure the temperature of the water; repeat this several times.

The advantage of the infrared thermometer is that it reduces the transfer of heat between the water and the thermometer.

The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made.

The equipment to use is: a beaker, a fixed amount of water, a thermometer.

The main precaution that must be taken is not to open the microwave while it is on.

4 0
2 years ago
what us the velocity in meters per second of a runner who runs exactly 110 m toward the beach in 72 secs
finlep [7]

Velocity = Distance / time taken

   =  110 m /72s

     1.52  ms⁻¹

4 0
3 years ago
Ted throws an object straight up into the air with an initial velocity of 54 ft/s from a platform that is 40 ft above the ground
elena-s [515]

Answer:

The time it will take for the object to hit the ground will be 4.

Explanation:

You have:

h(t)=−16t²+v0*t+h0

Being v0 the initial velocity (54 ft/s) and  h0 the initial height (40 ft) and replacing you get:

h(t)=−16t²+54*t+40

To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:

0=−16t²+54*t+40

Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values ​​of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:

\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}

In this case you have that:

  • a=-16
  • b= 54
  • c= 40

Replacing in the expression of the calculation of roots you get:

\frac{-54+\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}  Expresion (A)

and

\frac{-54-\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)} Expresion (B)

Solving the Expresion (A):

\frac{-54+\sqrt{5476 } }{2*(-16)}= \frac{-54+74}{2*(-16)}=\frac{20}{2*(-16)}=\frac{20}{-32}= -\frac{5}{8}

Solving the Expresion (B):

\frac{-54-\sqrt{5476 } }{2*(-16)}= \frac{-54-74}{2*(-16)}=\frac{-128}{2*(-16)}=\frac{-128}{-32}= 4

These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then <u><em>the time it will take for the object to hit the ground will be 4.</em></u>

6 0
2 years ago
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