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MrRissso [65]
3 years ago
13

Calculate the size of the labor force. The size of the adult population is 30,000. The number working is 17,700 and the number o

f unemployed actively looking for work is 1,600.
Physics
2 answers:
allsm [11]3 years ago
4 0

Answer:

The size of the labor force is 19300.

Explanation:

Given that,

Adult population = 30000

Number of  employed = 17700

Number of unemployed = 1600

Suppose, The labor force is comprised of those who are working for any number of hours (the employed) and those who are out of work but actively looking for a job the unemployed)

We need to calculate the size of the labor force

The labor force is comprised of number of employed person and number of unemployed person.

That means,

Labor Force = Employed + Unemployed

F_{l}=17700+1600

F_{l}=19300

Hence, The size of the labor force is 19300.

zhuklara [117]3 years ago
3 0

Answer:

Labor force = 19,300

Explanation:

Given that

Adult population = 30,000

Number of employed labor= 17,700

Number of unemployed labor= 1,600

The labor force is given as

Labor force =  Number of employed + Number of unemployed labor

Now by putting the values in the above equation we get

Labor force = 17,700 + 1,600

Labor force = 19,300

Therefore the labor force will be 19,300.

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In order to make an object stop, what direction does the unbalanced force need to be applied?
nataly862011 [7]
Against the object in motion
6 0
4 years ago
A supply plane drops a package to the ground for stranded campers below. After falling for 8.14 seconds, the package will have a
pshichka [43]

Answer:

333 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 m/s.

Time (t) = 8.14 s.

Final velocity (v) = 80.79 m/s.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

v² = u² + 2gh

80.79² = 0 + (2 × 9.8 × h)

6527.0241 = 0 + 19.6h

6527.0241 = 19.6h

Divide both side by 19.6

h = 6527.0241 / 19.6

h = 333 m

Thus, the plane was at a height of 333 m when the package was dropped.

5 0
3 years ago
Please help!!
xz_007 [3.2K]

Displacement is the area under the velocity/time graph. So for example this object's displacement in the first 3 seconds is (1/2)(3sec)(12.5 m/s)= 18.75m. (and then it starts backing up, displacement decreasing, after 3sec when velocity is negative).

But This object is never speeding up. Its velocity is smoothly decreasing at (25/6) m/s^2 (the slope of the graph). So the answer to the question is actually zero.

3 0
4 years ago
Suppose that two sources of sound produce waves of the same exact amplitude and frequency, but are out of phase by one-half of a
tia_tia [17]

Answer:

option D is right

........

8 0
3 years ago
Read 2 more answers
I can’t figure this out!!! Answer what you can , please.
tigry1 [53]

a/b. The ball has velocity vector at time t

\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)

where v_0=16\dfrac{\rm m}{\rm s} is the ball's initial speed and g=9.8\dfrac{\rm m}{\mathrm s^2}.

c. At its highest point, the ball has 0 vertical speed. This occurs when

v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s

d. Recall that

{v_y}^2-{v_{0y}}^2=-2g\Delta y

so that at its highest point,

0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m

e. This is just twice the time it takes for the ball to reach its maximum height, t=2.9\,\mathrm s.

f. The ball's horizontal position after time t is

v_0\cos63^\circ\,t

so that after the time found in part (f), the ball has traveled

v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m

4 0
4 years ago
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