Learning Task 1: Analyze the figure and answer the questions that follow.

1. Which one is the charged object?

2. What made the hair of the girl to rise?​

Answer:

T₂ = 16.83°C

Explanation:

Applying the law of conservation of energy principle here in this situation we get the following equation:

[tex]Energy\ Lost\ by\ Metal = Energy\ Gaine\ by\ Water\\m_{metal}C_{metal}(T_2-T_{1metal}) = m_{w}C_{w}(T_2-T_{1w})[/tex]

where,

T₂ = Final Temperature of Water = Final Temperature of Metal = ?

C_metal = Specififc Heat Capacity of the metal = 2.11 x 10² J/lg.°C

T_1metal = Initial Temperature of Metal = 225°C

m_metal = mass of metal = 1 x 10⁻²[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC) = (0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\2.11 T_2 - 474.75 = 376.56T_2 - 6778.08\\374.45T_2 = 6303.33\\[/tex] kg (exponent assumed due to missing info in question)

C_w = Specififc Heat Capacity of the water = 4184 J/lg.°C

T_1w = Initial Temperature of water = 18°C

m_w = mass of water = 0.09 kg

Therefore,

[tex](0.01\ kg)(211\ J/kg.^oC)(T_2-225^oC)=(0.09\ kg)(4184\ J/kg.^oC)(T_2-18^oC)\\\\2.11 - 474.75T_2 = 376.56 - 6778.08T_2\\[/tex]

T₂ = 16.83°C

Explanation:

Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.

So, we have:

P1/T1=P2/T2

Solving for P2, we have:

P2=(P1*T2)/T1

NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!

Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:

P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]

P2=228930.7156 N/m^2

Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2

Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!

Hope this helps!

Mark as brainlist...

Answer:

Friction.

Explanation:

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

Answer:  

A. α = - 1.047 rad/s²  

B. θ = 14.1 rad  

C. θ = 2.24 rev  

Explanation:  

A.  

We can use the first equation of motion to find the acceleration:

[tex]\omega_f = \omega_i + \alpha t[/tex]  

where,  

ωf = final angular speed = 0 rad/s  

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s  

t = time = 3 s  

α = angular acceleration = ?  

Therefore,

[tex]0\ rad/s = 3.14\ rad/s + \alpha(3\ s)[/tex]  

α = - 1.047 rad/s²

B.  

We can use the second equation of motion to find the angular distance:

[tex]\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2[/tex]  

θ = 14.1 rad

C.  

θ = (14.1 rad)(1 rev/2π rad)  

θ = 2.24 rev

Answer:

A. The amount of mass changes only slightly during a chemical

reaction.

Answer:

[tex]-180.38\ \text{N}[/tex]

Explanation:

[tex]q_1=-53\ \mu\text{C}[/tex]

[tex]q_2=105\ \mu\text{C}[/tex]

[tex]q_3=-88\ \mu\text{C}[/tex]

r = Distance between the charges

[tex]r_{12}=0.5\ \text{m}[/tex]

[tex]r_{23}=0.95\ \text{m}[/tex]

[tex]r_{13}=1.45\ \text{m}[/tex]

k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]

Net force is given by

[tex]F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}[/tex]

The force on the particle [tex]q_1[/tex] is [tex]-180.38\ \text{N}[/tex].

Answer:

The answer sir would be 180.38

Explanation:

Put in 180.38 trust

Answer:

are nearby and bright i hope this helps

Answer:

the focal length of the eyepiece is 17.23 cm

Explanation:

The computation of the focal length of the eyepiece is shown below:

= Focal length of objective lens ÷ angular magnification magnitude

= 86.3 ÷ -5.01

= 17.23 cm

Hence, the focal length of the eyepiece is 17.23 cm

We simply divided the angular magnification magnitude from the focal length of objective lens so that the focal length of the eyepiece could come

Answer:

strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.

Explanation:

The lines of magnetic force are drawn so that the density of lines is proportional to the intensity of the magnetic field.

Therefore, the sections where the magnetic field is strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.

Answer:

Homogeneous mixtures

Explanation:

I think so because homogeneous means mixed mixtures

This question is incomplete, the complete question is;

The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.

Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.

1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K

2) For this adiabatic expansion, what is the final temperature?  T[tex]_f[/tex] =  ? K

Answer:

1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2) the final temperature is 158.66 K

Explanation:

Given the data in the question;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

P[tex]_i[/tex] = 100 kPa = 100000 Pa

V[tex]_i[/tex] = 0.01 m³

V[tex]_f[/tex] = 0.026 m³

T[tex]_i[/tex] = 300 K

1)  the change in entropy due to the volume change alone

from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])

so change in entropy due to the volume change alone is;

ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])

we know that, from ideal gas law; PV = nRT

so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex]  ---- let this be equation 1

∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])

we substitute

ΔS = [( 100000 Pa ×  0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )

ΔS = 3.185 J/K  

Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.

2)  Final temperature

we know that, in an adiabatic expansion;

[tex]PV^Y[/tex] = K

where Y = 5/3

so

[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]

[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]

we substitute

[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]

[tex]P_f[/tex] = 20341.255 Pa

Also from ideal gas law;

PV = nRT

T = PV / nR

so

T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR

but from equation 1; nR = PV/T

so

T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )

T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )

we substitute

T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ ×  300 K) / 100000 Pa × 0.01 m³ )

T[tex]_f[/tex] = 158.66 K

Therefore, the final temperature is 158.66 K

Answer:

E = 3.6 x 10⁷ N/C

Explanation:

The electric field strength due to a point charge is given by the following formula:

[tex]E = \frac{kq}{r^2}\\\\[/tex]

where,

E = Electrical Field Strenght = ?

k = Colomb's Constant = 9 x 10⁹ Nm²/C²

q = magnitude of charge = 10 μC = 1 x 10⁻⁵ C

r = distance = 5 cm = 0.05 m

Therefore,

[tex]E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(1\ x\ 10^{-5}\ C)}{(0.05\ m)^2}[/tex]

E = 3.6 x 10⁷ N/C

Answer:

I think is is

Explanation:

B and C why because i have a gut feeling

Answer:

P = 3.92 10¹⁰ W

Explanation:

The power is data by the expression

         P = W / t

the work of a force is

         W = F. y

the bold ones represent vectors. In this case the displacement is vertical upwards and the vertical forces upwards, therefore the angle is zero and the cos 0 = 1

          W = F y

we substitute

          P = F y / t

          P = F v

as the body rises at constant speed the acceleration is zero and from the equilibrium condition

            F -W = 0

            F = mg

we substitute

           P = m g v

let's calculate

           P 1.00 10⁹ 9.8  4

           P = 3.92 10¹⁰ W

Answer:

[tex]193743.49\ \text{Pa}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Europa = [tex]4.74\times 10^{22}\ \text{kg}[/tex]

R = Radius of Europa = [tex]\dfrac{3130}{2}=1565\ \text{km}[/tex]

h = Depth = 150 m

[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]

Acceleration due to gravity is given by

[tex]g=\dfrac{GM}{R^2}[/tex]

Pressure is given by

[tex]P=\rho gh\\\Rightarrow P=\rho \dfrac{GM}{R^2}h\\\Rightarrow P=1000\times \dfrac{6.674\times 10^{-11}\times 4.74\times 10^{22}}{1565000^2}\times 150\\\Rightarrow P=193743.49\ \text{Pa}[/tex]

The submarine should be designed to withstand a pressure of [tex]193743.49\ \text{Pa}[/tex].

The pressure due to column of liquid increases with the gravitational

attraction, density of the liquid and height of the liquid column.

Reasons:

Mass of Europa, M = 4.78 × 10²² kg

Diameter of Europa, r = 3,130 km

Required:

The pressure a submarine must be designed to withstand when submerged to a depth of 150 meters.

Solution:

Gravitational acceleration on Europa is given by the formula;

[tex]\displaystyle g = \mathbf{\frac{G \cdot M}{r^2}}[/tex]

G = Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

[tex]Radius \ of \ the \ satellite\ Europa = \dfrac{3,130 \ km}{2} = \mathbf{1,565 \ km}[/tex]

[tex]\displaystyle g = \frac{6.67408 \times 10^{-11} \times 4.78 \times ^{22}}{1565000^2} \approx 1.303[/tex]

The gravitational acceleration on Europa, g ≈ 1.303 m/s²

Therefore;

Pressure due to a liquid's weight, P = ρ·g·h

The pressure at 150 m, P₁₅₀ = 150 × 1.303 × 997 = 194863.65

The pressure the submarine must be designed to withstand is, P₁₅₀ ≈ 194863.65 Pa.

Learn more here:

https://brainly.com/question/24337968

Answer:  The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.

Explanation:

Answer:

v = 2683.33 m/s

Explanation:

The magnetic force and the electric force on the electron must be the same, in order for them to cancel each other:

[tex]Electric\ Force = Magnetic\ Force\\Eq = qvBSin\theta \\\\v = \frac{E}{BSin\theta}[/tex]

where,

v = velcoity of electron = ?

E = Electric Field = 3059 V/m

B = Magnetic Field = 1.14 T

θ = Angle between velocity and magnetic field = 90°

Therefore,

[tex]v = \frac{3059\ V/m}{(1.14\ T)Sin90^o}[/tex]

v = 2683.33 m/s

The amount of heat absorbed by the water will be 8368 J.

Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.

It is given by the formula as ;

[tex]\rm Q= mcdt[/tex]

The given data in the problem is;

Equilibrium temperature = 30°C.

mass of water  = 50 g  ,

Temperature change = 70ºC

The specific heat of water =4.184 J//g °C

The amount of heat absorbed by the water is;

[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]

Hence, the amount of heat absorbed by the water will be 8368 J.

To learn more about the heat gain refer to the link;

https://brainly.com/question/26268921

#SPJ2

Answer:

the correct answue are  B, A, C, C, B

Explanation:

1) The electric field is requested, let's approximate the membrane by a parallel plate with surface charge density

         E = [tex]\frac{\sigma }{2 \epsilon_o }[/tex]

         E = [tex]\frac{ 10^{-5}}{2 \ 8.85 \ 10^{-12}}[/tex]

         E = 5.65 10⁵ N / C

the correct answer is B

2) A calcium ion has two positive charges, so the force applied by each side of the membrane (plate)

         F = q E

         F = 2  1.6 10⁻¹⁹  5.65 10⁵

         F = 1.8 10⁻¹³ N

the total force is the sum of the force of each membrane and the two forces go to the same side

         F = total = 2 F

         F_total = 3.6 10⁻¹³ N

the correct answer is A

3) the field and the electric potential are related

          ΔV = - E s

          ΔV = - 5.65 10⁵  10 10⁻⁹

          ΔV = - 5.65 10⁻³ V

          the correct answer is C

4) In the exercise they indicate that the outer wall has a positive charge, therefore, as they indicate that we approximate the system to a capacitor, the inner wall must be negatively charged.

The electric field goes from the positive to the negative charge, which is why it goes from the outer wall to the inner wall

the correct answer is C

5) For this part we use conservation of energy

starting point. On the inside wall, brown

            Em₀ = U = qV

final point. On the outside

             Em_f = K

energy is conserved

           Em₀ = Em_f

           q V = K

            K = 3 10⁻¹⁵  5.65 10⁻³

            K = 1.7 10⁻¹⁷ J

the correct answer is B

Answer:

what does the outer part of the disk turn into

Explanation:

4) it gets sucked into the star

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

[tex]T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s[/tex]

FOR THE WAVE TRAVELING THROUGH AIR:

[tex]T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s[/tex]

The separation in time between two pulses can now be given as follows:

[tex]\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\[/tex]

ΔT = 0.02412 s

The period of the vertical mass-spring is 0.64 s.

The angular speed of the vertical mass-spring is calculated as follows;

[tex]V_{max} = A \omega\\\\\omega = \frac{V_{max}}{A} \\\\\omega = \frac{7.02}{0.713} \\\\\omega = 9.85 \ rad/s[/tex]

The period of the vertical mass-spring is calculated as follows;

[tex]f = \frac{\omega }{2\pi} \\\\T = \frac{1}{f} \\\\T = \frac{2 \pi}{\omega } \\\\T = \frac{2\pi }{9.85} \\\\T = 0.64 \ s[/tex]

Thus, the period of the vertical mass-spring is 0.64 s.

Learn more about period of oscillation here: https://brainly.com/question/20070798

Answer:

Explanation:

The answer is c. I am very sure

Answer:

i think its b

Explanation:

im not very sure

its wave length

its wave lenght because how its measure

Answer:

Its Greater potential energy because the air is high up and that makes high energy power

Explanation: