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3 years ago

Among the elements potassium, lithium, and iron, the metallic bonds are likely to be strongest in

Physics

1 answer:

Mrac [35]3 years ago

4 0

彼が好きな男のような話は妻と関係があったのでそう

Explanation:これが難しい場合はコメントで教えてくださいが何でも

lol okay

from: a random person :)

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A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t

Zina [86]

Answer:

The value of spring constant is 266.01 $\frac{N}{m}$

Explanation:

Given:

Mass of pellet $m = 2.01 \times 10^{-2}$ kg

Height difference of pellet rise $h_{f} - h_{o} = 6.03$ m

Spring compression $x = 9.45 \times 10^{-2}$ m

From energy conservation law,

Spring potential energy is stored into potential energy,

$mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}$

Where $k =$ spring constant, $g = 9.8 \frac{m}{s^{2} }$

$k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }$

$k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }$

$k = 266.01$ $\frac{N}{m}$

Therefore, the value of spring constant is 266.01 $\frac{N}{m}$

6 0

3 years ago

Can someone help me with this activity?

Inessa05 [86]

What class is that in if math or biology I’m not good that

8 0

3 years ago

A pulley lifts a 72-N load with a force of 24-N. The input distance is 2m and the output distance is 0.5m. What is the efficienc

Debora [2.8K]

Answer:

Explanation:

Work done on the lever ( input energy ) = force applied x input distance

= 24 N x 2m = 48 J

Work done by the lever ( output energy ) = load x output distance

= 72 N x 0.5m = 36 J

efficiency = output energy / input energy

= 36 J / 48 J

= 3 / 4 = .75

In percentage terms efficiency = 75 % .

5 0

3 years ago

A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio

saveliy_v [14]

Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

6 0

3 years ago

mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago