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Zina [86]
3 years ago
5

What can you predict about an element from its position in the periodic table

Physics
1 answer:
bazaltina [42]3 years ago
5 0
Based on its position in the periodic table, you can predict how many electrons it has, how many valence electrons, how many levels of electrons, and its atomic number and mass.
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SCORPION-xisa [38]

Answer:

2. the volume of the square are the same

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2 years ago
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A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?
pychu [463]

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

H=ut+\frac{1}{2}gt^2

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m

3 0
3 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®
Furkat [3]

C. Kevlar

:) ++ * ++ !

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3 years ago
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What describes the particles in a liquid
Gnoma [55]

Answer:

liquid a particles slides past pother

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mark brainliest :))

8 0
3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
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