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Katen [24]
3 years ago
12

The more _________ an electron has, the further away it can be from the nucleus. mass

Physics
2 answers:
Monica [59]3 years ago
4 0

Answer:

ectron's further from the nucleus are held more weakly by the nucleus, and thus can be removed by spending less energy. Hence we say they have higher energy. The Coulomb interaction energy between a nucleus of atomic number and an ele

sasho [114]3 years ago
3 0

Answer: the more energy  

Explanation:

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A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat
umka21 [38]

Answer:

Av_2 =24\ V/V

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width  = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

                           Av₁ Bw₁ = Av₂ Bw₂

                           96 x 8 = Av₂ x 32

                           Av_2= \dfrac{96\times 8}{32}

                           Av_2 =24\ V/V

the highest gain available under these conditions Av_2 =24\ V/V

6 0
3 years ago
A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun
melomori [17]
Constant = straight line
“Travels at constant negative acc.”
Which is negative slope

Solution: B. Straight line w/ neg. slope
7 0
3 years ago
Carbohydrates, fats, proteins: These organic molecules ALL have this element in common. This element can form multiple bonds, an
statuscvo [17]
Carbon is the answer. IF oxygen were on the list it would also be correct but for this its Carbon<span />
6 0
3 years ago
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A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
A car travels a distance of 98 meters in 10 seconds. What is the average speed of the car?
stira [4]

Answer:

C

Explanation:

4 0
3 years ago
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