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gogolik [260]
3 years ago
6

In 2013, how many of the 800,000 black teenagers who participated in the labor market

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
310,400 black teenagers had participated in the labor market, hope this helps!
Alinara [238K]3 years ago
4 0
344,000 were unemployed
456,000 were employed 
614,000 would have been employed if they had the same unemployment rate as white teenagers
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What must a system owner do, once the threshold leak rate has been exceeded on any low-pressure system using an ozone-depleting
soldier1979 [14.2K]

Answer:

A procedure according to the norms.

Explanation:

If possible, proceed to fix the leak in no more than 30 days from the moment it was discovered.

Otherwise, during the first 30 days develop a planification to backfit the leak or, if needed, retire the appliance. This should be executed within one year.

7 0
3 years ago
A person hangs from a nylon rope (Young's modulus of 5 x 109 N/m2) as seen in the picture below. The rope stretches by 2 % and h
disa [49]

Answer:

959183.7 kg  

Explanation:

from the question we have :

young modulus = 5 x 10^{9} N/m^{2}

strain = 2% = 2÷100 = 0.02

diameter = 0.03 m

radius = 0.015 m

acceleration due to gravity (g) = 9.8 m/s^{2}

we can get the mass from the formula below

young modulus = stress ÷ strain

where

stress = \frac[force}{area} = \frac {mass x g}{area}

area = 2πr = 2π x 0.015 = 0.094

therefore    

young modulus = \frac{\frac {mass x g}{area}}{strain}

 5 x 10^{9}  =  \frac{\frac {mass x 9.8}{0.094}}{0.02}

mass =  \frac{5 x 10^{9} x 0.02 x 0.094}{9.8}

mass = 959183.7 kg  

8 0
3 years ago
Why is it difficult to use the law of conservation of energy to calculate the
Liula [17]

Answer:

Inelastic collisions actually do conserve energy, but the loss of energy to heat and mechanical vibration is hard to calculate so the math equating energy before and energy after is hard to balance.

7 0
3 years ago
Read 2 more answers
A 11 547,83-N car is raised using a hydraulic lift, which
kati45 [8]

Answer:

75457.54816 N

Explanation:

F_1 = Initial force

F_2 = Weight of car = 11547.83 N

r_1 = Smaller radius = 5 cm

r_2 = Larger radius = 19.56 cm

From Pascal's law we have

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_1=F_2\dfrac{A_1}{A_2}\\\Rightarrow F_1=11547.83\dfrac{\pi r_1^2}{\pi r_2^2}\\\Rightarrow F_1=11547.83\dfrac{5^2}{19.56^2}\\\Rightarrow F_1=754.57548\ N

The force is 754.57548 N

3 0
4 years ago
How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius
mixas84 [53]

The magnitude of static friction is

<em>f</em> = <em>mv</em> ²/<em>r</em>

(i.e. the net force acting on the car parallel to the road points toward the center of the curve)

while the net vertical force must be

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

because the car is otherwise in equilibrium. Then

==>   <em>n</em> = <em>mg</em>

==>   <em>f</em> = <em>µn</em> = <em>µmg</em> = <em>mv</em> ²/<em>r</em>

==>   <em>µ</em> = <em>v</em> ²/(<em>rg</em>)

We have

<em>v</em> = 101 km/h ≈ 28.1 m/s

<em>r</em> = 110 m

<em>g</em> = 9.80 m/s²

so that

<em>µ</em> = (28.1 m/s)² / ((110 m) <em>g</em>) ≈ 0.730

6 0
3 years ago
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