10 You are given the following list of vo-
lumes: 0.45 m3; 375 cm3; 75 cL; 0.6 dm3.
a. Express all the volumes of the list in
liters.
b. Arrange these volumes in a descending
order.

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:

what does the outer part of the disk turn into

Explanation:

4) it gets sucked into the star

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:

fffffgggggggggggggghhh

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

Answer:

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

Explanation:

Answer:

Homogeneous mixtures

Explanation:

I think so because homogeneous means mixed mixtures

its wave length

its wave lenght because how its measure

Answer:

Explanation:

The answer is c. I am very sure

Answer:

i think its b

Explanation:

im not very sure

Wouldn't it be B because it's a majority pointing to it?

Sorry if i'm wrong.

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

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:

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) The total resistance of the circuit is 25 Ω

(B) The current through each resistor is 4.4 A

(C) For 10Ω: Potential drop = 44 V

     For 15Ω: Potential drop = 66 V

Explanation:

Given;

potential difference, V = 110V

resistors in series, = 10Ω and a 15Ω

(A) The total resistance of the circuit is calculated as follows;

Rt = 10Ω + 15Ω = 25Ω

(B) The current through each resistor;

Same current will flow through the two resistors since they are in series.

I = V/Rt

I = 110 / 25

I = 4.4 A

(C) The voltage drop across each resistor;

For 10Ω: Potential drop = IR₁ = 4.4 x 10 = 44 V

For 15Ω: Potential drop = IR₂ = 4.4 x 15 = 66 V

The magnitude of the angular acceleration of the pendulum at the moment of release is; α = 18.45 rad/s²

We are given;

Mass of rod; m = 1 kg

Length of rod; L = 49.8 cm = 0.498 m

Mass of Disk; M = 4 kg

Radius of disk; r = 24.9 cm = 0.249 m

Let us first calculate the torque acting from the formula;

τ = mg(L/2) + MgL

Thus;

τ = (1 × 9.8 × (0.498/2)) + (4 × 9.8 × 0.498)

τ = 21.96 N.m

Using parallel axis theorem, we can find the moment of inertia about the given axis as;

I = (mL²/3) + ½MR² + ML²

Plugging in the relevant values gives;

I = (1 * 0.498²/3) + ½(4 * 0.249²) + (4 * 0.498²)

I = 1.19 kg.m²

The angular acceleration is given by the formula;

α = I/τ

α = 21.96/1.19

α = 18.45 rad/s²

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The unit called horsepower is defined based on the amount of work a horse can do in 1 minute.

James Watt used the horsepower unit for the first time in 1782. According to the account, James Watt used a pony to charge coal from the mines. According to the narrative, he needed a unit to measure the force of one of these animals.

He discovers that they can move 22.000 lbs per minute and decides to (arbitrarily) boost this metric by 50%. Having been the unit 33.000 lb/ft per minute, horsepower is an engine's output horsepower rating, whereas brake horsepower is an engine's input brake horsepower.

Brake horsepower measures an engine's power without accounting for power losses, whereas HP accounts for power losses. Horsepower of the brakes

Therefore, the horsepower unit is defined by the amount of work a horse can do in one minute.

To learn more about horsepower, refer to the link:

https://brainly.com/question/8652948

#SPJ2

Answer:

Virtual

Explanation:

Answer:

B. Virtual

Good Luck!

Answer:

correct answer is A

Explanation:

The diagram shows a series circuit with three resistors and two power sources.

In a series circuit the current through the entire circuit is constant and the resistance is the sum of the resistances in the circuit.

When the power sources are placed in opposite position the voltage between them is subtracted.

     V_b - 30 = I (10 + 4 + 6)

     V_b = I (20) - 30

      V_b = 30 - 2 20

       V_b = 10 V

the correct answer is A

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:

the distance is 315.3696 m

Explanation:

The computation of the distance is given below:

Given that

Sound intensity = 1.67 × 10^-6 W/m^2

And, the distance = 233 m

Now as we know that

Power = Intensity × surface area

1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2

d^2 = 2 × (223)^2

= √2 × 223

= 315.3696 m

Hence, the distance is 315.3696 m

Answer:

Yes, the value will be the same.

Explanation:

Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.

Answer:

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

reaction.

Answer:

manage your weight better, have stronger bones, have lower blood pressure, less risk of a heart attack, etc.

Answer:

You become healthier, your body starts regulating better, you get stronger bones and muscles, and you lower the risk of diabetes,heart problems and other diseases.

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:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   [tex]\frac{m}{m+ \frac{I}{r^2} }[/tex]

          v² = 2gh    [tex]\frac{1}{1 + \frac{I}{m r^2} }[/tex]

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = [tex]g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }[/tex]

let's use trigonometry

         sin θ = h / L

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    [tex]\frac{1}{1 + \frac{mr^2 }{m r^2 } }[/tex]1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  [tex]\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }[/tex] = g sin θ   [tex]\frac{1}{1+ \frac{1}{2} }[/tex]

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   [tex]\frac{1}{1 + \frac{2}{3} }[/tex]

     a₃ = g sin θ [tex]\frac{3}{5}[/tex]

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  [tex]\frac{1 }{1 + \frac{2}{5} }[/tex]

     a₄ = g sin θ  [tex]\frac{5}{7}[/tex]

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      [tex]\frac{105}{210}[/tex]

b) a₂ = g sinθ ⅔ = g sin θ     [tex]\frac{140}{210}[/tex]

c) a₃ = g sin θ [tex]\frac{3}{5}[/tex]= g sin θ       [tex]\frac{126}{210}[/tex]

d) a₄ = g sin θ [tex]\frac{5}{7}[/tex] = g sin θ      [tex]\frac{150}{210}[/tex]

the order of acceleration from lower to higher is

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

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.

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