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1 year ago

11

While looking at calcium (Ca) on the periodic table, a student needs to find an element with a greater atomic mass in the same p

eriod. Where should the student look?

2 answers:

inessss [21]1 year ago

7 0

Directly to the right :)

stiks02 [169]1 year ago

5 0

To the right of calcium in the same period.

Hope this helps!

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The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

density is $10^{6}$ Mg/µL

Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to $10^{6}$ cm³

and here 1 cm³ is equal to 1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = $10^{18}$ kg/m³

density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

density = $10^{6}$ Mg/µL

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1 year ago

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A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward

Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

Area of trapezoidal are=

= $\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)$

= $\frac{1}{2} *(2.25-1.75*0.5)$

=0.6875 m

As the area is positive therefore displacement from xo is positive

ax=(change in velocity)/(Time)

ax= $\frac{-0.5-0}{3} =-\frac{1}{6} ms2$

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

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1 year ago

A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J

faltersainse [42]

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

Final K = 1/2 Iω²

W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

Thus, the final angular velocity is 20 rad/s .

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1 year ago

Light hits a clear plastic bottle and a granite rock. Which choice most accurately describes the effect of visible light on the

balu736 [363]

Answer:

#4 is the accurate answer.

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2 years ago

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