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2 years ago

12

Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

n

2 answers:

egoroff_w [7]2 years ago

6 0

Explanation:

Chlorine is the smallest atomic radius

Kaylis [27]2 years ago

4 0

Chlorine is the smallest because up right extreme of the periodic table

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Verizon [17]

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● Hypothesis
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3 years ago

What is amplitude?

nydimaria [60]

1. Frequency is the number of complete waves that pass a point in a second. 2.Wavelength is the distance between two crests or two troughs. 3.Time period <span> is the time it takes for one complete wave to pass a given point. 4. Amplitude is the height of the wave. Hence option 4 is correct. </span>

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2 years ago

At a speed of 60 mph, how far does your car travel in 3 seconds

Viefleur [7K]

One of the many random useless factoids that I carry around
in my head is the factoid that 60 miles per hour is equivalent
to exactly 88 feet per second.

So in three seconds at that speed, you would cover exactly

(3 x 88) = 264 feet.

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2 years ago

Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's<br> displacement?

tresset_1 [31]

Answer:

Joes displacement is 54m

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1 year ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

2 years ago