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STatiana [176]
3 years ago
12

What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The eleme

nts have the same number of electrons. C. The properties change going across each row. D. The elements in each row combine easily.
Physics
1 answer:
Sliva [168]3 years ago
7 0
The right answer for the question that is being asked and shown above is that: "C. <span>. The properties change going across each row. " the </span>statement that applies to the horizontal rows or periods in the periodic table is that t<span>he properties change going across each row. </span>
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A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the
Svetllana [295]

Answer:

a) \Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}, b) \Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}, c) \Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}

Explanation:

a) The net flux through the cube is:

\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}

b) The flux through the right face is:

\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}

c) The flux through the left face is:

\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}

5 0
3 years ago
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Would an astronauts mass change as she traveled from planet to planet? Explain.
Lunna [17]
No, it's mass remains the same. It's attraction to the planets center AKA gravity changes.
7 0
3 years ago
Computer-enhanced x-rays are called __________ scans.<br> A. TMS<br> B. PET<br> C. CT<br> D. MRI
goldenfox [79]

Answer:

C. CT

Explanation:

It stands for Computed Tomography Scans.

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3 years ago
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An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
dsp73

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

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3 years ago
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Answer:

During the segments B - C and D - E, the car stopped since the y axis is the distance and the distance stayed the same in between those segments.

For a simpler answer, the flat horizontal lines on the graph are the times when the car was stopped.

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