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spin [16.1K]
3 years ago
14

Which is not an affect of alcohol

Engineering
1 answer:
Gre4nikov [31]3 years ago
6 0
Pooping problems is not an affect
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Which of the following is most useful for doing research?
Ghella [55]

Answer:

Web Browser

Explanation:

Because you dont use a messaging app or presentation software to look up stuff its common knowledge

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2 years ago
A TPMS (tire pressure monitoring system) instrument panel indicator lamp is on. Technician A says the most likely cause is low t
zheka24 [161]

Answer:

Actualmente estoy trabajando en una pregunta diferente en este momento.

Explanation:

Actualmente estoy trabajando en una pregunta diferente en este momento.

6 0
3 years ago
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc
a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
The steel bracket is used to connect the ends of two cables. if the allowable normal stress for the steel is sallow = 30 ksi, de
garri49 [273]

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

<h3>The static equilibrium is given as:</h3>

F = P (Normal force)

Formula for moment at section

M = P(4 + 1.5/2)

= 4.75p

Solve for the cross sectional area

Area = \frac{\pi d^{2} }{4}

d = 1.5

\frac{\pi *1.5^{2} }{4}

= 1.767 inches²

<h3>Solve for inertia</h3>

\frac{\pi *0.75^4}{4}

= 0.2485inches⁴

Solve for the tensile force from here

\frac{F}{A} +\frac{Mc}{I}

30x10³ = \frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\

30000 = 14.902 p

divide through by 14.902

2013.15 = P

The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

Read more on tensile force here: brainly.com/question/25748369

4 0
2 years ago
The path taken to move items from their origin to their destination is known as which of the following?
mrs_skeptik [129]

Answer:d

Explanation:

7 0
2 years ago
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