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3 years ago

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If the demand function for a commodity is given by the equation p2 + 16q = 1200 and the supply function is given by the equation

300 − p2 + 2q = 0, find the equilibrium quantity and equilibrium price. (Round your answers to two decimal places.)

1 answer:

Svetradugi [14.3K]3 years ago

5 0

Answer:

p = 20, q = 50

Explanation:

p² + 16q = 1200

300 - p² + 2q = 0

By adding the two equations, we can eliminate p²:

16q + 300 + 2q = 1200

18q = 900

q = 50

Solving for p:

p² + 16(50) = 1200

p = 20

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Which of the following statements describes an electric generator?

belka [17]

An electric generator is when a magnet is rotated through a coil of wire to produce an electric current.

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In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
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6 0

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

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3 years ago

A 100-N force causes an object to<br> accelerate at 2 m/s/s. What is the<br> mass of the object?

Kitty [74]

Answer:

$50\; \rm kg$ .

Explanation:

By Newton's Second Law, the acceleration $a$ of an object is proportional to the net force $\sum F$ on it. In particular, if the mass of the object is $m$ , then

$\sum F = m \cdot a$ .

Rewrite this equation to obtain:

$\displaystyle m = \frac{\sum F}{a}$ .

In this case, the assumption is that the $100\; \rm N$ force is the only force that is acting on the object. Hence, the net force $\sum F$ on the object would also be

Make sure that all values are in their standard units. Forces should be in Newtons (same as $\rm kg \cdot m \cdot s^{-2}$ , and the acceleration of the object should be in meters-per-second-squared ( $\rm m \cdot s^{-2}$ ). Apply the equation $\displaystyle m = \frac{\sum F}{a}$ to find the mass of the object.

$\displaystyle m = \frac{100\; \rm N}{2\; \rm m \cdot s^{-2}} = 50\; \rm kg$ .

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3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

3 years ago