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3 years ago

Ls -2 a solution of 4x +3= -5?.

Physics

1 answer:

torisob [31]3 years ago

5 0

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

x=−2

Hope it helps,

Please mark me as the brainliest

Thank you

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Use the equation d=m/v. If a rock has a density of 2g/cm^3 and a volume of 8cm^3, what is the mass?

Ray Of Light [21]

Answer:

The mass is 16 [g]

Explanation:

We have to remember the formula that defines density, which mentions that the density is equal to mass divided by the volume.

$Density = mass / volume\\where:\\density = 2[\frac{g}{cm^{3} } ]\\Volume = 8 [cm^{3}]$

Now we clear the mass:

$mass= density*volume\\mass = 2[\frac{g}{cm^{3} }]*8[cm^{3} ]\\mass = 16 [g]$

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3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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3 years ago

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A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

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3 years ago

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PSYCHO15rus [73]

Explanation:

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3 years ago

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ankoles [38]

Answer:

Ok look Im going try my best, and the answer that most likely looks like it can be correct is C the third picture i hope this is correct and helps bye bye:3

also do you wanna get some points >w>

Explanation:

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3 years ago

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Ls -2 a solution of 4x +3= -5?.​

Ls -2 a solution of 4x +3= -5?.