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igor_vitrenko [27]
2 years ago
14

Which material(s) listed below is an example of a persistent organic pollutant?

Physics
1 answer:
grigory [225]2 years ago
4 0

Answer:

mercury

arsenic

ricin

ddt

Explanation:

pls mark me as brainliest

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Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.
zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

heat= m s \Delta T


m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= 20(s_1)(70-30)=800s_1


Heat gained by fluid=100(s_2)(30-20)=1000s_2


Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0
3 years ago
Plz do all of it i will give brainlest and thanks to best answer<br> plz do it right
AlladinOne [14]
The answer is a rainforest I’m pretty sure
3 0
3 years ago
Read 2 more answers
Choose the three statements that are true about valence electrons.
Varvara68 [4.7K]

We have that valence electrons poses the three characteristics stated, as

Group 14 (carbon group)  are identified by 4 valence electrons.

Valence electrons of atoms are used to form bonds.

Group 14 (carbon group)  are identified by 4 valence electrons.

Option A,B,C

<h3>Properties of Valence electrons</h3>

All elements in the same group or family have the same number of valence electrons: Yes, this is true as Group 14 (carbon group) are identified by 4 valence electrons.

Valence electrons are the only subatomic particles involved in forming bonds: Yes, Valence electrons of atoms are used to form bonds.

Carbon has 4 valence electrons because it is found in group 14:

True, Group 14 (carbon group)  are identified by 4 valence electrons.

For more information on atoms visit

brainly.com/question/13981855

6 0
2 years ago
A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir
Viefleur [7K]

Answer:

y = 10.44cos(2t - 0.291) cm

Explanation:

y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)

v = y' = -2Αsin(2t + φ)

10 = Acos(2(0) + φ) = Acosφ

6 = -2Αsin(2(0) + φ) = -2Asinφ

6/10 = -2Asinφ/Acosφ = -2tanφ

tanφ = -0.3

φ = -0.291 radians

10 = Acos(-0.291)

A = 10/cos(-0.291) = 10.44

7 0
3 years ago
Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l
ziro4ka [17]

Answer:

The pressure is 6570  lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

a_1 = \sqrt{\gamma RT } =  \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7}  = 3

then, M_n = M_1sin \beta = 3sin(35) = 1.721

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570  lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

M_t_1 = M_1cos \beta = 3 cos(35) = 2.458

w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s

a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s

v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s

Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7}  \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o

6 0
3 years ago
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