All categories

3 years ago

Please help i'm going to throw up from stress

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

$t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})}$ where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

$16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)}$ and

$16.0=\frac{83640x+2928}{4182x+292.8}$ and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

You might be interested in

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

Which of the following describes resistance force?

Verdich [7]

Force applied by the machine to over come resistance

5 0

3 years ago

Read 2 more answers

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

MATHPHYS HELP

rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0

3 years ago

calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm

ZanzabumX [31]

Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)

8 0

3 years ago