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murzikaleks [220]

2 years ago

4

An ideal gas occupies a volume of 0.60 m3 at 5.0 atm and 400. K. What volume does it occupy at 4.0 atm and a temperature of 200.

K?
What is the known and unknown when using P1V1/T1=P2V2/T2

1 answer:

Wewaii [24]2 years ago

3 0

It is given that P1=5atm
V1=0,6m3
T1=400K
And
P2=4atm
T2=200K
Therefore the unknown is V2
From now on you put all the known in the formula P1xV1/T1=P2xV2/T2 and with the help of algebra find what the unknown is ☺️

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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

garik1379 [7]

Answer:

T=2.94*10^-10 N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm

(λ= wavelength of standing wave; as there will be no node

except the endpoints of silk strand)

λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

150 * 0.28 = √(T/μ) ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

= [π * (10 * 10^(-6))²] * 1300 = 13π * 10^(-8).

now putting this in equation (#) we get

150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)² =

2.94*10^-10 N/m.

6 0

2 years ago

Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may displ

olga2289 [7]

Answer:

What is u should know it bc u should answered it already

Explanation:

7 0

2 years ago

Read 2 more answers

The gravitational force between two asteroids is 6.2 × 108 n. asteroid y has three times the mass of asteroid z. if the distance

SOVA2 [1]

Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.

Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)

The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)

Answer: 1.1x10¹⁶ kg

4 0

2 years ago

Read 2 more answers

The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10–15 s–1 at 298 K and a rate constant of 8

netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

$ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})$

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

$ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})$

Solving for Ea as:

Ea = 159982.23 J /mol

1 J/mol = 10⁻³ kJ/mol

Ea = 159.98 kJ/mol

7 0

2 years ago

A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an

kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

$m_Hv_H + m_Av_A = m_Hv_2+m_Av_2$

From the question given, we can deduce that the anvil is at rest;

∴ $v_A = 0$ ; then, we have:

$m_Hv_H + 0 = (m_H+m_A) v_2$

Making $v_2$ the subject of the formula; we have:

$v_2 = \frac{m_Hv_H}{m_H + m_A}$ ------- Equation (1)

Also, from the second diagram; there is a representation of a free body momentum of the hammer head;

From the diagram;

F = impulsive force exerted on the rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

$m_Hv_H - F \delta t = m_Hv_2$

$- F \delta t = - m_Hv_H + m_Hv_2$

$F \delta t = m_Hv_H - m_Hv_2$

$F \delta t = m_H(v_H - v_2)$ ------- Equation (2)

Using the function of the kinetic energy of the hammer before impact $T_1$ ; we have:

$T_1 = \frac{1}{2} m_Hv_H^2$ -------- Equation (3)

We determine the mass of the hammer $m_H$ by using the formula from:

$W_H = m_Hg$

where;

$W_H$ = weight of the hammer

$m_H$ = mass of the hammer

g = acceleration due to gravity

Making $m_H$ the subject of the formula; we have:

$m_H = \frac{W_H}{g}$

$m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}$

$m_H = 0.04658 \ lb.s^2/ft$

Now;

$T_1 = \frac{1}{2} m_Hv_H^2$

$T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2$

$T_1 = \frac{18.632 }{2}$

$T_1 = 9.316 \ ft.lb$

After the impact $T_2$ ; the final kinetic energy of the hammer and anvil can be written as:

$T_2 = \frac{1}{2}(m_H +m_A)v^2_2$

Recall from equation (1) ; where $v_2 = (\frac{m_Hv_H}{m_H+m_A})$ ; if we slot that into the above equation; we have:

$T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2$

$T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}$

$T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})$

Also; from equation (3)

$T_1 = \frac{1}{2} m_Hv_H^2$ ; Therefore;

$T_2 = T_1 (\frac{m_H}{m_H+m_A})$ ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil $m_A = \frac{W_A}{g}$

where we replace; $W_A \ with \ \infty$ and g = 32.2 ft/s²

$m_A = \frac{\infty}{32.2 \ ft/s}$

However ; from equation (1)

$v_2 = \frac{m_H v_H}{m_H + m_A}$

$v_2 = \frac{0.04658*20}{0.04658+ \ \infty}$

$v_2 = 0$

From equation (2)

$F \delta t = m_H(v_H + v_2)$

$F \delta t = (0.04658 lb .s^2 /ft )(20ft/s - 0)$

$F \delta t = \ 0.932 \ lb.s$

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

For the energy absorbed by the rivet ; we have:

$T_2 = T_1 (\frac{m_H}{m_H+m_A} )$

where;

$T_1= 9.316 \ ft.lb$

$m_H = 0.04658 \ lb.s^2/ft$

$m_A = \infty$

Then;

$T_2 = (9.316 \ ft.lb) (\frac{0.04658\ lb.s^2/ft)}{0.04658 \ lb.s^2/ft+ \infty} )$

$T_2 = (9.316 \ ft.lb)* 0$

$T_2 = 0$

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = $T_1 - T_2$

ΔT = 9.316 ft.lb - 0

ΔT ≅ 9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b)

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer for the part b of this question.

The attached file can be found below.

5 0

2 years ago