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zepelin [54]
3 years ago
10

A bungee jumper starts with 1000 J in their GPE store. After they jump they fall and are brought to a stop with the bungee cord.

With the cord fully stretched what is the energy in the EPE store?
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

energy is equal to 1000 J

Explanation:

When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J

This is a case of energy transformation, but the total value of mechanical energy does not change

         

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A. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.
vovikov84 [41]

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

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3 years ago
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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that
Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

T = \frac{135\ s}{98}

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

T = 2\pi \sqrt{\frac{l}{g}}

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}

<u>g = 11.2 m/s²</u>

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3 years ago
All heat engines involve _____. the refrigeration cycle evaporation and condensation combustion
pishuonlain [190]
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4 0
3 years ago
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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}.

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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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