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3 years ago

Which type of transformer is found outside a residential house

1 answer:

7 0

The type of transformer that is found outside a residential house is :
step-down transformer
In step-down transformer, the secondary voltage is less than the primary voltage, which will reduce the voltage from the primary win.

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A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

Allisa [31]

$1.6a = \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3} \\ \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\ \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\ \\ m_3 = \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\ \\ m_3 = 22.57 kg$

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3 years ago

The brainstem is to breathing and arousal as the limbic system is to memories and: (2 points)

riadik2000 [5.3K]

Answer:

The brainstem is to breathing and arousal as the limbic system is to memories and emotion. See the explanation below, please.

Explanation:

The limbic system develops certain responses to emotional stimuli (joy, anger, sadness, fear, pleasure). It interacts with the autonomic endocrine and nervous system. It is also related to sexuality, memory, attention, memories. It includes some areas such as hypothalamus, tonsil, hippocampus, among others.

8 0

3 years ago

A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0

ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

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

Which formula can be used to solve problems related to the first law of thermodynamics?

docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

8 0

3 years ago

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