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3 years ago

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A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t

ime between the person see and hear a firework explosion

1 answer:

WINSTONCH [101]3 years ago

3 0

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

$v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s$

So, the required time is 2.2 seconds.

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Explanation:

It is given that,

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or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

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3 years ago

A student drops a 1.5 kg rock off of a 4.0 m tall bridge. The speed of the rock just before it hits the water below is 6 m/s. Ho

olga nikolaevna [1]

Answer:

Energy due to air resistance = 31.8 Joules

Explanation:

According to the law of conservation of energy, energy can neither be created nor destroyed but can be transformed from one form to another

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If there is no energy loss due to air resistance, potential energy = kinetic energy

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3 years ago

Read 2 more answers

A wave has a wavelength of 7 mm and a frequency of 9 hertz. What is its speed?

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Speed of wave = Frequency x Wavelength
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3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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3 years ago