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3 years ago

Match the correct term with the definition

Physics

2 answers:

prohojiy [21]3 years ago

7 0

Hey there!!
see answer in attachment.

ki77a [65]3 years ago

4 0

1. Atomic Mass - protons + neutrons in the nucleus of an atom

2. Neutron - the neutrally charged element in the nucleus of an atom

3. Atomic Number - the number of protons an atom has

4. Proton - the positive particle in the nucleus of an atom

5. Isotope - an atom of an element that has a different number of neutrons that the standard atom of that element

Hope This Helps!

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Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

Paraphin [41]

Answer:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

Explanation:

The 4 wires are connected in series: this means that the same current flow through them, and the voltage of the battery, V0, is equal to the sum of the voltages on each individual resistor:

$V_0=V_1+V_2+V_3+V_4$

Also, the equivalent resistance of the series circuit is

$R_{eq}=R_1+R_2+R_3+R_4$

The voltage V2 across wire 2 is given by Ohm's law:

$V_2 = R_2 I$ (1)

where I is the total current in the circuit, which is given by:

$I=\frac{V_0}{R_{eq}}=\frac{V_0}{R_1+R_2+R_3+R_4}$

Substituting this into eq. (1), we find an expression for V2:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

8 0

3 years ago

A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0

3 years ago

The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere

spin [16.1K]

Answer:

P = 0.0644 atm

Explanation:

Given that,

The pressure of a sample of gas is measured as 49 torr.

We need to convert this temperature to atmosphere.

The relation between torr and atmosphere is as follow :

1 atm = 760 torr

1 torr = (1/760) atm

49 torr = (49/760) atm

= 0.0644 atm

Hence, the presssure of the sample of gas is equal to 0.0644 atm.

5 0

2 years ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$