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Zolol [24]
3 years ago
9

What's the opposite of futility​

Physics
1 answer:
pantera1 [17]3 years ago
6 0

Answer:

pointlessness or uselessness

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An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an
Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

Put the value into the formula

L'=\dfrac{4}{1.01953}

L'=3.92\ m

Hence, The length of the tube is 3.92 m.

8 0
3 years ago
Breaking a pencil in half is an example of: options:
Licemer1 [7]

Answer:

b

Explanation:

8 0
3 years ago
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Which landscape will erode more quickly? Assume the same precipitation and average temperature in each landscape.
Artemon [7]

Answer:

  B.  Landscape B

Explanation:

Shale is fine sediment pressed together to form rock.

Sandstone is larger (sand-grain-sized) sediment cemented together to form rock.

Shale erodes faster, as evidenced by the second attachment. That attachment shows erosion of a rock face consisting of interbedded shale and sandstone. The shale has receded significantly, leaving the sandstone layers with space between them.

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3 years ago
How do the presidential roles of chief executive and commander in chief differ?
Mashcka [7]
A. The commander in chief role deals only with the the military, while the chief executive role is broader
3 0
3 years ago
A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other
Soloha48 [4]

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

P_1 = 174kPA

\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg

P_2 = 175kPa \rightarrow Saturated vapor

\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg

P_e = 175kPa \rightarrow Saturated vapor

V = 4L

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2

Similarly through the energy balance you can get that

E_{in}-E_{out} = \Delta E_{system}

Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow Since W=ke=pe=0

The initial mass, initial internal energy, and final mass in the tank are

m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}

m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}

m_1 = 1.893+0.002= 1.895Kg

At the same time the internal energy can be defined from the mass in state 1 as,

U_1 = m_1 u_1

U_1 = m_fu_f+m_gu_g

U_1 = 1.893*486.82+0.002*2524.5

U_1 = 926.6kJ

The calculation of mass in state 2 can be defined as

m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}

m_2 = 0.004Kg

Then from the mass and energy balances,

m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg

In this way the calculation of the heat of entry would be subject to

Q_{in} = m_eh_e+m_2u_2-m_1u_1

Q_{in} = 1.891*2700.2+0.004*2524.5-926.6

Q_{in} = 4188kJ

Therefore the Power would be given as

\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW

Therefore the highest rate of heat transfer allowed is 1.163kW

3 0
3 years ago
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