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3 years ago

Please help only if you answer all 4

Physics

2 answers:

bonufazy [111]3 years ago

8 0

5.) decomposing leaves
6.) pretty sure is killer whale
7.) ecosystem
4.) D

Oduvanchick [21]3 years ago

5 0

Answer:

i think ice burg for 2nd page for penguin sorry if its wrong!

Explanation:

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What kind of engine do most cars use today

Nadusha1986 [10]

Answer:

4 stroke 4 cylinder Natural Aspirated Gasoline engine

Explanation:

im not 100% sure

4 0

3 years ago

Read 2 more answers

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

Number one, it's probably a really easy question but it confused me

Dvinal [7]

Use your notes or look for the answer on your book

6 0

4 years ago

Which of the following correctly identifies the initial direction of the force on the moving positively charged particle (i.e.,

meriva

Answer:

Explanation:

Given

Current is flowing towards right side

Considering right side i.e. +x to be positive and direction outside the paper be +k

suppose charge is below the wire and is moving towards right

$v=v_0\hat{i}$

magnetic field below the wire is outside the Paper using right hand thumb rule

$B=B_0\hat{k}$

$F=q(\vec{v}\times \vec{B})$

$F=q(v_0\hat{i}\times B_0\hat{k})$

$F=-qv_0B_0\hat{j}$

i.e. Force acts in Downward direction

i charge is moving towards left i.e.

$v=-v_0\hat{i}$

$F=q(-v_0\hat{i}\times B_0\hat{k})$

$F=qv_0B_0\hat{j}$ towards upward direction

6 0

3 years ago

The electric field of a charge is defined by the force on: An electron A probe charge A proton. A source charge.

Alina [70]

Answer:

A probe charge

Explanation:

As we know that electric field intensity is the force experienced by the probe charge which is placed in the electric field region

Here we can say it as

$E = \frac{F}{q}$

so here that probe charge should be very small so that it will not disturb the electric field in the space.

If the probe charge is of large magnitude then the field will get disturbed and the intensity which is to be measured is different from its actual value.

Also the sign of the probe charge is taken to be positive.

so correct answer here will be

A probe charge

7 0

3 years ago