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3 years ago

The distribution of ground shaking around the fault

1 answer:

6 0

Ground shaking is a term used to describe the vibration of the ground during an earthquake. Ground shaking is caused by body waves and surface waves. As a generalization, the severity of ground shaking increases as magnitude increases and decreases as distance from the causative fault increases.

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A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is

Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0

3 years ago

90% of traffic crashes are due to driver error. True False

kumpel [21]

90% of traffic crashes are due to driver error.

True

3 0

2 years ago

Read 2 more answers

Consider the following relational database that Best Airlines uses to keep track of its mechanics, their skills, and their airpo

Musya8 [376]

Answer:

Explanation:

SELECT MECHNAME,AGE FROM MECHANIC;

SELECT AIRNAME,SIZE FROM AIRPORT WHERE SIZE>=20 AND STATE='CALIFORNIA' AND YEAROPENED >=1935 ORDER BY SIZE ASC;

SELECT AIRNAME,SIZE FROM AIRPORT WHERE (SIZE>=20 OR YEAROPENED >=1935) AND STATE='CALIFORNIA';

SELECT AVG(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

SELECT COUNT(AIRNAME) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAROPENED>=1935 GROUP BY STATE;

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAR OPENED>=1935 GROUP BY STATE HAVING COUNT(*)>=5;

SELECT MECHNAME FROM MECHANIC A JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME AND B.STATE='CALIFORNIA';

SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILLNAME='FAN BLADE RELACEMENT';

J) SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILL NAME='FAN BLADE REPLACEMENT'

JOIN AIRPORT D

ON A.AIRNAME=D.AIRNAME

AND STATE='CALIFORNIA';

K) SELECT SUM(SALARY),CITY FROM MECHANIC A

JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME

AND STATE='CALIFORNIA'

GROUP BY CITY;

L) SELECT MAX(SIZE) FROM AIRPORT ;

M) SELECT MAX(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA';

6 0

3 years ago

The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along

JulsSmile [24]

Answer:

D=41.48 ft

$a=54.43\ ft/s^2$

Explanation:

Given that

y=0.5 x²

Vx= 2 t

We know that

$V_x=\dfrac{dx}{dt}$

At t= 0 ,x=0

$x=\int V_x.dt$

At t= 3 s

$x=\int_{0}^{3} 2t.dt$

$x=[t^2\left\right ]_0^3$

x= 9 ft

When x= 9 ft then

y= 0.5 x 9² ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

$D=\sqrt{9^2+40.5^2} \ ft$

D=41.48 ft

$a_x=\dfrac{dV_x}{dt}$

Vx= 2 t

$a_x= 2\ ft/s^2$

At t= 3 s , x= 9 ft

y=0.5 x²

$a_y=\dfrac{d^2y}{dt^2}$

y=0.5 x²

$\dfrac{dy}{dt}=x\dfrac{dx}{dt}$

$\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}$

Given that

$\dfrac{dx}{dt}=2t$

$\dfrac{dx}{dt}=2\times 3$

$\dfrac{dx}{dt}=6\ ft/s$

$a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2$

$a_y=54\ ft/s^2$

$a=\sqrt{a_x^2+a_y^2}\ ft/s^2$

$a=\sqrt{2^2+54^2}\ ft/s^2$

$a=54.43\ ft/s^2$

7 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)

dQ = mdq

dq = $C_p$ dT

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (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