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4 years ago

8

If the labor force of 155 million people is growing by 1.2 percent per year, how many new jobs have to be created each month to

keep unemployment from increasing?

2 answers:

ololo11 [35]4 years ago

7 0

1.2 percent of 155 million is 1,860,000. Divide that by 12 (for twelve months of the year) and get 155,000. That's the answer. 155,000.

amm18124 years ago

3 0

Answer:0.155 million jobs have to be created each month to keep unemployment from increasing.

Explanation:

Population of the labor force = 155 million

Percentage increase of labor force in an year = 1.2 % of 155 million

Percentage increase in labor force in a month = $\frac{1.2 \%}{12}=0.1 \%$ of 155 million

Jobs required to be created each month = population increase each month

$=155 million\times \frac{0.1}{100}=0.155 million$

0.155 million jobs have to be created each month to keep unemployment from increasing.

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The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving

Blizzard [7]

Answer:

$2.4\cdot 10^{-11} N$

Explanation:

Since the Earth's magnetic field is perpendicular to your direction of motion, the strength of the magnetic force exerted on your head is given by:

$F=qvB$

where:

$q=6\cdot 10^{-9}C$ is the charge on your head

$v=80 m/s$ is the speed at which you are moving

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field of the Earth

By substituting these numbers into the equation, we find the strength of the magnetic force:

$F=(6\cdot 10^{-9}C)(80 m/s)(5\cdot 10^{-5} T)=2.4\cdot 10^{-11} N$

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3 years ago

A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

geniusboy [140]

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

5 0

3 years ago

The chemical formula for glucose is C6H12O6. Therefore, four molecules of glucose will have carbon atoms, hydrogen atoms, and ox

Gekata [30.6K]

If there are 4 molecules of glucose, there will be 24 carbon, 48 hydrogen, and 24 oxygen.

5 0

3 years ago

Read 2 more answers

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t

lana [24]

Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

$q_{h}$ = R* $T_{h}$ in( $V_{2}$ / $V_{1}$ )

$=0.287 * 1200 ln(3)$

$=0.287*1318.33$

$=378.36kJ/kg$

The specific heat capacity:

$q_{L}$ =q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

$n_{TH}$ =1 - ( $T_{L}$ / $T_{H}$ )

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

4 0

3 years ago

Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou

spin [16.1K]

Answer:

(a) 28690 m/s (b) $2.46x10^{33}J$

Explanation:

The orbital speed is define as:

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

$v = \frac{d}{t}$ (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

$d = vt$ (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

$8min . \frac{60s}{1min}$ ⇒ $480s$

$t = 480s$

Replacing all those values in equation 3 it is gotten:

$d = (3.0x10^{8}m/s)(480s)$

$d = 1.44x10^{11}m$

Kepler’s third law is defined as:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

$T = \sqrt{r^{3}}$

$T = \sqrt{(1.44x10^{11}m)^{3}}$

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

$T = \sqrt{1AU}$

$T = 1AU$

That can be expressed in units of years.

$T = 1AU . \frac{1year}{1AU}$

$T = 1year$

But there are 31536000 seconds in one year:

$T = 1year . \frac{31536000s}{1year}$

$T = 31536000s$

Finally, equation 1 can be used:

$v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}$

$v = 28690 m/s$

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

$E = \frac{1}{2}mv^{2}$ (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

$F = \frac{GMm}{r^{2}}$

$ma = \frac{GMm}{r^{2}}$ (5)

M will be isolated in equation 5:

$M = \frac{r^{2}a}{G}$

Where r is the radius of the Earth ( $6.38x10^{6}m$ )

$M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})$

$M = 5.98x10^{24} Kg$

$E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}$

$E = 2.46x10^{33}Kg.m^{2}/s^{2}$

$E = 2.46x10^{33}J$

<u>Hence, the kinetic energy of Earth is $2.46x10^{33}J$ .</u>

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3 years ago