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3 years ago

The reactivity of an element is based on its __________.

Physics

1 answer:

Goryan [66]3 years ago

7 0

Answer:

The number of electrons in the outermost shell of an atom determines its reactivity. Noble gases have low reactivity because they have full electron shells. Halogens are highly reactive because they readily gain an electron to fill their outermost shell.

Explanation:

i think

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A spring attached to the ceiling is stretched one foot by a four pound weight. The mass is set in motion by pulling down 2 feet

Marina CMI [18]

Given that,

Weight = 4 pound

$W=4\ lb$

Stretch = 2 feet

Let the force be F.

The elongation of the spring after the mass attached is

$x=2-1=1\ feet$

(a). We need to calculate the value of spring constant

Using Hooke's law

$F=kx$

$k=\dfrac{F}{x}$

Where, F = force

k = spring constant

x = elongation

Put the value into the formula

$k=\dfrac{4}{1}$

$k=4$

(b). We need to calculate the mass

Using the formula

$F=mg$

$m=\dfrac{F}{g}$

Where, F = force

g = acceleration due to gravity

Put the value into the formula

$m=\dfrac{4}{32}$

$m=\dfrac{1}{8}\ lb$

We need to calculate the natural frequency

Using formula of natural frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, k = spring constant

m = mass

Put the value into the formula

$\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}$

$\omega=\sqrt{32}$

$\omega=4\sqrt{2}$

(c). We need to write the differential equation

Using differential equation

$m\dfrac{d^2x}{dt^2}+kx=0$

Put the value in the equation

$\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0$

$\dfrac{d^2x}{dt^2}+32x=0$

(d). We need to find the solution for the position

Using auxiliary equation

$m^2+32=0$

$m=\pm i\sqrt{32}$

We know that,

The general equation is

$x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})$

Using initial conditions

(I). $x(0)=2$

Then, $x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})$

Put the value in equation

$2=A+0$

$A=2$ .....(I)

Now, on differentiating of general equation

$x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})$

Using condition

(II). $x'(0)=0$

Then, $x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})$

Put the value in the equation

$0=0+\sqrt{32}B$

So, B = 0

Now, put the value in general equation from equation (I) and (II)

So, The general solution is

$x(t)=2\cos\sqrt{32t}$

(e). We need to calculate the time

Using formula of time

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4\sqrt{2}}$

$T=1.11\ sec$

Hence, (a). The value of spring constant is 4.

(b). The natural frequency is 4√2.

(c). The differential equation is $\dfrac{d^2x}{dt^2}+32x=0$

(d). The solution for the position is $x(t)=2\cos\sqrt{32t}$

(e). The time period is 1.11 sec.

5 0

3 years ago

If an object is dropped from a height of 144 feet, the function h(t)= -16t^2+144 gives the height of the object after t seconds.

kirill [66]

The equation given in the question is
<span>h(t)= -16t^2+144
When the height becomes zero, then
- 16t^2 + 144 = 0
16t^2 = 144
(4t)^2 = (12)^2
4t = 12
t = 3
From the above deduction, it can be easily concluded that the correct option among all the options that are given in the question is the fourth option or option "D". I hope that this is the answer that has actually come to your desired help.</span>

6 0

2 years ago

. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110

7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x $10^{5}$ J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, $v_{0}$ = 110 km/h = 110 x 1000/3600 = 30.6 m/s $\frac{30.6^{2} }{2x9.8}$

initial height, $h_{0}$ = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = $\frac{1}{2} mv_{0} ^{2}$

gh = $\frac{1}{2} v_{0} ^{2}$

h = $\frac{v_{0} ^{2} }{2g}$

= 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

= mg (h - $h_{0}$ )

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x $10^{5}$ J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

= 1.88 x $10^{5}$ /(22/sin 2.5°)

= 373 N

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3 years ago

What does a cell division allow all multicellular organisms to do

Levart [38]

I believe that the answer should be B. It makes the most sense to me.

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3 years ago

A feather is dropped on the moon from a height of 1.40 meters. The acceleration due to gravity on the moon is 1.67 m/s2. Determi

RoseWind [281]

Distance = (1/2) (acceleration) (time)²

1.4m = (0.835 m/s²) (time)²

(time)² = (1.4/0.835) s²

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3 years ago