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Citrus2011 [14]
3 years ago
6

Describe the possible long term impact of a catastrophic event

Physics
1 answer:
denpristay [2]3 years ago
4 0
However, the Northridge earthquake itself, a significant catastrophic event<span>, had ... prices, reflecting their increased </span>potential<span> for physical destruction. ... local neighborhoods and find no </span>long-term impact<span> on home values, resident incomes or occupancy rates. ... We briefly </span>describe<span> the variety of data sources used in the paper.</span>
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The resistance, R, to electricity of a cylindrical-shaped wire is given by the equation , where p represents the resistivity of
valentina_108 [34]

Answer:

infinity

Explanation:

Given that

Resistance = R

Resistivity = ρ

Length = L

Diameter = d

The resistance of wire R given as

R=\rho\dfrac{L}{A}

A=Area

A=\dfrac{\pi d^2}{4}

Now by putting the value of A

R=\rho\dfrac{L}{\dfrac{\pi d^2}{4}}

R=\rho\dfrac{4L}{\pi d^2}

When d tends to infinity then d² will also tends to infinity.

So when d tends to zero then the resistance tends to infinity.

Therefore answer is ---

infinity

7 0
3 years ago
Read 2 more answers
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
Electricity flows from what to what
DerKrebs [107]
Electricity flows from positive to negative
5 0
3 years ago
Read 2 more answers
This person is fishing in a small, freshwater body of water. This body of water flows into a larger body of water called a river
ANEK [815]
Since the small body of water is connected to a larger body of water. so the small body of water will than be called a stream
6 0
4 years ago
Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

Proven

5 0
3 years ago
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