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3 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul

1 answer:

7 0

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the <em>kinetic energy</em> in the system <em>must</em> <em>decrease</em>.

In fact, this isn't even a "result". The kinetic energy has to decrease <em><u>before</u></em> the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

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The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to each corner, a

Kipish [7]

Answer:

$-2.60097\times 10^{-6}\ C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance between charges = 2 cm

The electric force is given by

$F=\dfrac{k4\times 10^{-6}q_A}{r^2}$

Also

$F=\dfrac{k4\times 10^{-6}q_B}{r^2}$

The sum of the force is given by

$\sum F=\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30+\dfrac{k4\times 10^{-6}q_B}{r^2}cos 30\\\Rightarrow \sum F=2\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30\\\Rightarrow q_a=\dfrac{\sum Fr^2}{2\times k4\times 10^{-6}cos30}\\\Rightarrow q_a=\dfrac{405\times 0.02^2}{2\times 8.99\times 10^9\times 4\times 10^{-6}cos30}\\\Rightarrow q_a=2.60097\times 10^{-6}\ C$

Thus $q_a=q_b=-2.60097\times 10^{-6}\ C$ .

The negative sign is because the force points downward which is taken as negative

7 0

4 years ago

If the slowest instruction in the SCA executes in 12.5 ns, then what is maximum system clock frequency in MHz

Margaret [11]

Since the slowest instruction in the SCA executes in 12.5 ns, the maximum system clock frequency is 80 MHz

To answer the question, we need to know what frequency is.

<h3>What is frequency?</h3>

Frequency is the number of oscillations per second of a wave.

It is given by f = 1/T where T = period of wave

Now, given that the slowest instruction in the SCA executes in t = 12.5 ns, we need to calculate maximum system clock frequency, f.

<h3>What is the maximum system clock frequency?</h3>

So, f = 1/t

= 1/12.5 ns

= 1/(12.5 × 10⁻⁹ s)

= 1/12.5 × 10⁹ Hz

= 0.08 × 10⁹ Hz

= 80 × 10⁻³ × 10⁹ Hz

= 80 × 10⁶ Hz

= 80 MHz

So, the maximum system clock frequency is 80 MHz

Learn more about maximum system clock frequency here:

brainly.com/question/14636488

#SPJ11

5 0

2 years ago

A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the

Viktor [21]

Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s

4 0

3 years ago

Tonya is thinking about the topic presented in the text, "Do opposites really attract?" Which of her thoughts is an example of c

tigry1 [53]

tanya is dumb j j j j j j j j j jj j j j

6 0

4 years ago

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.