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3 years ago

Lucas plugs a lamp into the wall. The light bulb in the lamp turns on. What kind of energy transformation is this

2 answers:

7 0

A light bulb converts electric energy into heat and light energy, but it is different when you are using solar energy.Not that different though, the solar energy turns into electric energy that converts into heat and light energy.

pantera1 [17]3 years ago

5 0

Answer:

Electric energy from the outlet transforms to radiant energy and thermal energy in the lightbulb.

Explanation:

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In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit

arlik [135]

Answer:

Equilibrium quantity = 5

Equilibrium price = 40

Explanation:

given:

p = -x²-3x+80

p = 7x+5

For the equilibrium quantity the price from both the functions will be equal

thus, we have

-x² - 3x + 80 = 7x+5

⇒ x² +3x + 7x + 5 - 80 = 0

⇒x² + 10x - 75 = 0

now solving for x

x²- 5x + 15x -75 = 0

x(x-5) + 15(x-5) = 0

therefore, the two roots of the equation are

x = 5 and x = -15

since the quantity cannot be in negative

therefore, the equilibrium quantity will be = 5

now the equilibrium price can be found out by substituting the equilibrium quantity in any of the equation

thus,

p = -(5)² -3(5) + 80 = 40

p = 7(5) + 5 = 40

4 0

2 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$