A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J

Answer:

i₀ = V / R_i

Explanation:

For this exercise we use Ohm's law

         V = i R

          i = V / R

the equivalent resistance for

         [tex]\frac{1}{R_{eq}}[/tex] =  ∑ [tex]\frac{1}{R_i}[/tex]

if all the bulbs have the same resistance, there are N bulbs

         [tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]

         R_{eq} = R_i / N

we substitute

         i = N V / Ri

where i is the total current that passes through the parallel, the current in a branch is

         i₀ = i / N

         i₀ = V / R_i

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

Explanation:

700n I think friend .. if worng

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=[tex]\rho[/tex]

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

[tex]Density=\frac{mass}{volume}[/tex]

[tex]Mass=Density\times volume=Constant[/tex]

Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]

Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]

Using the formula

[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]

[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]

[tex]\rho_2=8\rho[/tex]

Hence, the density of  the compressed sphere=[tex]8\rho[/tex]

Option a is correct.

Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ [tex]F_{net}=mg-F_r[/tex]

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ [tex]F_{net} = ma[/tex]

To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,

⇒ [tex]ma=mg-F_r[/tex]

⇒    [tex]a=g-\frac{F_r}{m}[/tex]

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

Answer:

Both the ball takes equal time to reach to the ground.  

Explanation:

Two balls of same diameter

Let the height is h.

Mass of ball 1 is more than the mass of ball 2.

The second equation of motion is

[tex]h = u t +0.5 gt^2[/tex]

Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.

So, both the ball takes equal time to reach to the ground.  

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Answer:

7.9 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]

Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:

[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]

Plug everything in:

7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]

(1590kg the initial weight plus the weight of the added passenger)

Answer:

I think it would be (-7 C )..

Answer and Explanation:

a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.

b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.

Answer:

1.67 m

Explanation:

The potential energy change of the small ball ΔU equals the work done by the buoyant force, W

ΔU = -W

Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5

ΔU = mgΔh

ΔU = ρVgΔh

Now, W = ρ'VgΔh'   where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h

So, ΔU = -W

ρVgΔh = -ρ'VgΔh'

ρVg(h - 5) = -ρ'Vgh

ρ(h - 5) = -ρ'h

Since the density of the small ball equals 1/2 the density of water,

ρ = ρ'/2

ρ(h - 5) = -ρ'h

(ρ'/2)(h - 5) = -ρ'h

ρ'(h - 5)/2 = -ρ'h

(h - 5)/2 = -h

h - 5 = -2h

h + 2h = 5

3h = 5

h = 5/3

h = 1.67 m

So, the maximum depth the ball reaches is 1.67 m.

Answer:

Static friction depends on the mass of the object.

Explanation:

Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;

1) nature of the object and the surface(how rough or smooth the surfaces are)

2)surface area of the object and the surface

3) mass of the object

Since;

F=μmg

Where;

μ= coefficient of static friction

m= mass of the object

g= acceleration due to gravity

Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.

Answer:

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Explanation:

GIven;

Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m

Diameter =  1.19 mm = 1.19 x 10⁻³ m

Find:

Minimum angular spread (in rad)

Computation:

Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]

Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]

Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]

Electric field, [tex]E=55\ N/C[/tex]

We need to find the distance. We know that, the electric field a distance r is as follows :

[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]

So, the required distance is 20.22 m.

Answer:

2m

Explanation:

wavelength=speed/frequency

=340/170

=2m

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Answer:

θ = 57.4°

Explanation:

The complete formula to find out the work done by the plane is as follows:

[tex]W = FdCos\theta[/tex]

where,

W = Work = 200000 J

F = Force = Tension = 2560 N

d = distance = 145 m

θ = angle between rope and horizontal = ?

Therefore,

[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]

θ = 57.4°

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Answer:

Hope you find it useful. please correct me if I am wrong

The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.

Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.

Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.

Given the force due to the weight of the elevator = mg = 2400N

m = 2400/9.8 Kg

The elevator deaccelerating while moving upward, a = -1.7 m/s²

According to Newton's 3rd law: T - mg = ma

T - 2400 = (2400/9.8) × (-1.7)

T = 2400 - 416.32

T = 1983.67 N

Learn more about tension, here:

brainly.com/question/28965515

#SPJ5

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

Answer:

Young's modulus for the rope material is 20.8 MPa.    

Explanation:

The Young's modulus is given by:

[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

[tex] F = T_{w} = W_{o} [/tex]

Where:

[tex] T_{w} [/tex]: is the tension of the wire

[tex]W_{o} [/tex]: is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]

Hence, the Young's modulus is:

[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]          

Therefore, Young's modulus for the rope material is 20.8 MPa.                

I hope it helps you!                                    

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]

Velocity acquired during this time

[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]

Consider vertical motion

[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]

Net velocity is

[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]

Angle made is

[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Answer: 2.4 km/hr

Explanation:

Distance = 600m

Time= 15 minutes = 15 x 60 second/minute = 900 seconds

Speed =  [tex]\frac{Distance}{Time}[/tex]  =  [tex]\frac{600}{900}[/tex]  = [tex]\frac{2}{3}[/tex] m/sec

⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)

The object's kinetic energy changes according to

dK/dt = 15 J/s

If v is the object's initial speed, then its initial kinetic energy is

K (0) = 1/2 (5 kg) v ²

Use the fundamental theorem of calculus to solve for K as a function of time t :

[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]

After t = 13 s, the object's kinetic energy is

K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J

Put this as the left side in the equation above for K(t) and solve for v :

[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]

==>   v ≈ 9.5 m/s

Answer:

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

Answer:

false.

Explanation:

Ok, we define average velocity as the sum of the initial and final velocity divided by two.

Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

AV = (V + (-V))/2  = 0

While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.

So we already can see that the average velocity will not be equal to half of the average speed.

The statement is false