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Vika [28.1K]
3 years ago
14

Jennifer, who has a mass of 48.0 kg, is riding at 40.2 m/s in her red sports car when she must suddenly slam on the brakes. An a

ir bag deploys in the car. She strikes the air bag, bringing her body to a stop in 0.56 s. What average force does the air bag exert on her
Physics
2 answers:
andre [41]3 years ago
6 0

Answer:

F = 3,445.71 N

Explanation:

Finding the deceleration (a):

a = \frac{u - v}{t}

initial velocity = u = 40.2 m/s

final velocity = v = 0 m/s

a = 40.2/0.56 = 71.785 m/s^2

Now, Finding the force:

F = ma

mass = m = 48.0 kg

a = 71.785 m/s^2

the force is found to be 3,445.71 N

Bogdan [553]3 years ago
3 0

Answer:

F = 3445.714\,N

Explanation:

The system can be modelled appropriately by the use of the Principle of Momentum Conservation and Impact Theorem:

(48\,kg)\cdot (40.2\,\frac{m}{s} ) - F\cdot (0.56\,s) = 0\,\frac{kg\cdot m}{s}

The average force exerted on her:

F = 3445.714\,N

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D

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Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

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h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
Identify the 3 types of energy transfer shown in the diagram.
fgiga [73]

Answer:

1.Conduction.

2.Convection.

3.Radiation.

Explanation:

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7 0
2 years ago
An object moves along the x axis. The graph shows it’s position x as a function of time t. Find the average velocity of the obje
nadezda [96]
Question is from B to C

Answer: (b) 1.5m/s

x1=3m, x2=9m
t1=1s, t2=5s
Displacement, ∆x=(9-3)m=6m
Time elapsed, ∆t=(5-1)s=4s

So average velocity v =∆x/∆t=6/4=1.5m/s
6 0
3 years ago
What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Marianna [84]

Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)

\Delta P=1.58\times10^{6}\ N/m^2

(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

8 0
3 years ago
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