An object is oscillating on a spring with a period of 4.60 s. At time t = 0.00 s the object has zero speed and is at x = 8.30 cm. What is the acceleration of the object at t = 2.50 s?

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder =  150 mm

thickness of cylinder = 6 mm

maximum shear stress =  25 MPa

we know that

hoop stress is given as =[tex]\frac{pd}{2t}[/tex]

axial stress is given as =[tex]\frac{pd}{4t}[/tex]

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

[tex]25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}[/tex]

[tex]25 = \frac{pd}{8t}[/tex]

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

Answer:

The power dissipated in the 3 Ω resistor is P= 5.3watts.

Explanation:

After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.

The resultating resistor is of Req=6Ω.

I= V/Req

I= 2A

the parallel resistors have a potential drop of Vparallel=4 volts.

I(3Ω) = Vparallel/R(3Ω)

I(3Ω)= 1.33A

P= I(3Ω)² * R(3Ω)

P= 5.3 Watts

Answer:

The magnitude of the car's centripetal acceleration is ac= 0.816 m/s² .

Explanation:

r= 1500m

Vt= 35 m/s

ac= Vt²/r

ac= (35m/s)² / 1500m

ac=0.816 m/s²

Answer:

   Option B is the correct answer.

Explanation:

We have resistance [tex]R=\frac{\rho l}{A}[/tex]

The resistance of wire B is four times that of wire A

        [tex]R_B=4R_A\\\\\frac{\rho_B l_B}{A_B}=4\times \frac{\rho_A l_A}{A_A}\\\\\frac{A_A}{A_B}=4\\\\\frac{\pi r_A^2}{\pi r_B^2}=4\\\\\frac{r_A}{r_B}=2[/tex]

    Option B is the correct answer.          

Answer:

462 nm

Explanation:

Given: width of the slit, d = 5.6 × 10⁻⁴ m

Distance of the screen, D = 4.0 m

Fringe width, β = 3.3 mm = 3.3 × 10⁻³ m

First dark fringe means n =1

Wavelength of the light, λ = ?

[tex] \beta = \frac{\lambda D}{d}\\ \Rightarrow \lambda = \frac{d \beta}{D} =\frac{5.6\times 10^{-4} \times 3.3 \times 10^{-3}}{4.0} = 4.62 \times 10^{-7}m = 462 nm[/tex]

Answer:

Time of race  = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

         s = ut + 0.5at²

         17 = 0 x t + 0.5 x 3.89 x t²

          t = 2.96 s

Velocity after 2.96 seconds

          v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

          [tex]t=\frac{83}{11.50}=7.22s[/tex]

Time of race = 2.96 + 7.22 = 10.18 s

Answer:16.096

Explanation:

Given

mass of dog[tex]\left ( m_d\right )=10kg[/tex]

mass of boat[tex]\left ( m_b\right )=46kg[/tex]

distance moved by dog relative to ground=[tex]x_d[/tex]

distance moved by boat relative to ground=[tex]x_b[/tex]

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0=[tex]m_d\times x_d+m_b\dot x_b[/tex]

0=[tex]10\times x_d+46\dot x_b[/tex]

[tex]x_d=-4.6x_b[/tex]

i.e. boat will move opposite to the direction of dog

Now

[tex]|x_d|+|x_b|=7.8[/tex]

substituting[tex] x_d [/tex]value

[tex]5.6|x_b|=7.8[/tex]

[tex]|x_b|=1.392m[/tex]

[tex]|x_d|=6.4032m[/tex]

now the dog is  22.5-6.403=16.096m from shore

The distance from the shore after the dog walks on the boat is calculated using the conservation of momentum. Since there are no external forces, the center of mass of the dog-boat system must stay the same. After the dog walks towards the shore, we determine the boat's movement in the opposite direction and calculate the dog's final distance from the shore.

To solve how far the dog is from the shore after walking on the boat, we need to apply the law of conservation of momentum. Since there's no external force acting on the system (the dog plus the boat), the center of mass of the system will not move. Initially, the center of mass is stationary, and it should remain so as the dog walks.

We can calculate the initial position of the center of mass (CM) using the formula:

CM = (m × dog_position + M × boat_center_position) / (m + M)

As the dog walks 7.8 m towards the shore, the boat moves in the opposite direction to keep the center of mass of the system in the same place. Let's denote the distance that the boat moves as x. Using the conservation of momentum:

m × dog_walk = M × x

The distance from the shore after the dog walks = initial dog's distance from shore - dog's walk + boat's movement towards shore (which is x).

However, since we're keeping the center of mass stationary, boat's movement towards shore (x) is equal to the dog's walk distance (7.8 m) multiplied by the ratio of the dog's mass to the boat's mass.

Therefore:

x = (m/M) × dog_walk

And so, the final distance from the shore = 22.5m - 7.8m + x

Answer:

Option C is the correct answer.

Explanation:

Young's modulus is the ratio of tensile stress and tensile strain.

Bulk modulus is the ratio of pressure and volume strain.

Rigidity modulus is the ratio of shear stress and shear strain.

Here we are asked about Young's modulus which is the ratio of tensile stress and tensile strain.

Option C is the correct answer.

Answer:

C.

It is the force per unit area acting on the material’s surface.

Explanation:

Answer:

2.06 x 10^-7 s

Explanation:

For going opposite to the applied electric field

u = 6.2 x 10^5 m/s, v = 0 ,  q = 1.6 x 19^-19 C, E = 62000 N/C,

m = 1.67 x 106-27 Kg, t1 = ?

Let a be the acceleration.

a = q E / m = (1.6 x 10^-19 x 62000) / (1.67 x 10^-27) = 6 x 10^12 m/s^2

Use first equation of motion

v = u + a t1

0 = 6.2 x 10^5 - 6 x 10^12 x t1

t1 = 1.03 x 10^-7 s

Let the distance travelled is s.

Use third equation of motion

V^2 = u^2 - 2 a s

0 = (6.2 x 10^5)^2 - 2 x 6 x 10^12 x s

s = 0.032 m

Now when the proton moves in the same direction of electric field.

Let time taken be t2

use second equation of motion

S = u t + 1/2 a t^2

0.032 = 0 + 1/2 x 6 x 10^12 x t2^2

t2 = 1.03 x 10^-7 s

total time taken = t = t1 + t2

t =  1.03 x 10^-7 +  1.03 x 10^-7 =  2.06 x 10^-7 s

Answer:

4.08 x 10⁻¹⁵ m

Explanation:

r₀ = constant of proportionality = 1.3 x 10⁻¹⁵ m

r = radius of the nucleus 31P = ?

A = mass number of the nucleus 31P = 31

Radius of the nucleus is given as

[tex]r=r_{o}A^{\frac{1}{3}}[/tex]

Inserting the values

[tex]r=(1.3\times 10^{-15})(31)^{\frac{1}{3}}[/tex]

[tex]r=(1.3\times 10^{-15})(3.14)[/tex]

r = 4.08 x 10⁻¹⁵ m

Answer:

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Explanation:

The perigee and apogee of the satellite are 1.275 x 10^4 km and 2.125 x 10^4 km, respectively.

The perigee and apogee of a satellite can be determined using the equations:

Perigee = Semimajor Axis - Eccentricity x Semimajor Axis

Apogee = Semimajor Axis + Eccentricity x Semimajor Axis

Given that the semimajor axis is 1.7 x 10^7 m and the eccentricity is 0.25, we can substitute these values into the equations to find:

Perigee = (1.7 x 10^7 m) - (0.25 x 1.7 x 10^7 m) = 1.275 x 10^7 m

Apogee = (1.7 x 10^7 m) + (0.25 x 1.7 x 10^7 m) = 2.125 x 10^7 m

Converting these values to km:

Perigee = 1.275 x 10^4 km

Apogee = 2.125 x 10^4 km

Answer:

6593.4

Explanation:

wavelength, λ = 910 nm = 910 x 10^-9 m

Speed of laser = 3 x 10^8 m/s

t = 20 pico seconds

Distance traveled by laser in this time, d = c x t

d = 3 x 10^8 x 20 x 10^-12 = 6 x 10^-3 m

number of wavelengths, n = d / λ

n = (6 x 10^-3) / (910 x 10^-9)

n = 6593.4

Answer:

6.44 × 10^10 N/C

Explanation:

Electric field due to the ring on its axis is given by

E = K q r / (r^2 + x^2)^3/2

Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.

r = 0.25 m, x = 0.5 m, q = 5 C

K = 9 × 10^9 Nm^2/C^2

E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2

E = 6.44 × 10^10 N/C

Answer:

299.51 m/s

Explanation:

m = mass of the bullet = 45 g = 0.045 kg

M = mass of the block = 1.55 kg

v = muzzle speed of the bullet

V = speed of bullet-block combination after the collision

μ = Coefficient of friction between the block and the surface = 0.28

d = distance traveled by the block = 13 m

V' = final speed of the bullet-block combination = 0 m/s

acceleration of the bullet-block combination due to frictional force is given as

a = - μg

using the kinematics equation

V'² = V² + 2 a d

0² = V² + 2 (- μg) d

0 = V² - 2 (μg) d

0 = V² - 2 (0.28) (9.8) (13)

V = 8.45 m/s

Using conservation of momentum for collision between bullet and block

mv = (M + m) V

(0.045) v = (1.55 + 0.045) (8.45)

v = 299.51 m/s

Answer:

F = 0.7 N

Explanation:

As we know that the wire touch the ends of the diagonal of the rectangle

so here the length of the wire is given as

[tex]L = \sqrt{x^2 + y^2}[/tex]

[tex]L = \sqrt{0.6^2 + 0.8^2}[/tex]

[tex]L = 1 m[/tex]

now force due to magnetic field on current carrying wire is given as

[tex]F = iLB[/tex]

now we have

[tex]F = (7)(1)(0.1)[/tex]

[tex]F = 0.7 N[/tex]

Answer:

g = 8.82 m/s/s

It is somewhat less than 9.81 m/s2.

Explanation:

As we know that the formula to find the gravitational acceleration is given as

[tex]g = \frac{GM}{r^2}[/tex]

here we know that

r = distance from center of earth

here we have

[tex]r = R + h[/tex]

[tex]r = (6370 + 350) km = 6720 km [/tex]

now we have

[tex]g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6720 \times 10^3)^2}[/tex]

[tex]g = 8.82 m/s^2[/tex]

Answer:

The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]

Explanation:

Given that,

Radius = 2.2 cm

Charge [tex]q = 6.08\times10^{-6}\ C[/tex]

Angular speed = 2.01 rad/s

We need to calculate the time period

[tex]T = \dfrac{2\pi}{\omega}[/tex]

Now, The spinning produced the current

Using formula for current

[tex] I = \dfrac{q}{T}[/tex]

We need to calculate the magnetic moment

Using formula of magnetic moment

[tex]M = I A[/tex]

Put the value of I and A into the formula

[tex]M=\dfrac{q}{T}\times A[/tex]

[tex]M=\dfrac{q\times\omega}{2\pi}\times \pi\times r^2[/tex]

Put the value into the formula

[tex]M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}[/tex]

[tex]M=2.96\times10^{-9}\ Am^2[/tex]

Hence, The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]

The magnetic moment of the rotating ring with a radius of 2.2 cm, a total charge of 6.08 µC, and an angular velocity of 2.01 rad/s is approximately 1.176 x 10⁻⁵ A⋅m².

The magnitude of the magnetic moment (μ) of a rotating charged ring can be found using the formula given by the product of the charge (Q) and the angular velocity (ω) and the radius (r) squared, as μ = Qωr²/2. For a uniform non-conducting ring rotating about an axis perpendicular to its plane, this relationship can be used to determine the magnetic moment generated by its rotation.

To calculate the magnetic moment for the given scenario:

Carrying out the multiplication gives the magnetic moment:

μ = (6.08 x 10-6 C)(2.01 rad/s)(0.000484 m2)/2 = 0.01176 x 10-6 C⋅m2/s

This results in a magnetic moment of approximately μ ≈ 1.176 x 10-5 A⋅m2.

Using a formula linking initial and final angular velocities, angular acceleration, and total angle, the final angular velocity can be obtained. Multiplying the final angular velocity and radius gives the final linear velocity. Angular acceleration and linear acceleration on the rim of a wheel are related by their radius.

Given the angular acceleration, a, the number of revolutions, and the radius of the wheel, r, you can determine the angular velocity at the end of 2 revolutions using the formula w_f = sqrt(w_i^2 + 2*a*θ), where w_i is the initial angular velocity, w_f is the final angular velocity, a is angular acceleration, and θ is the total angle swept out in radians. The angular velocity can then be used to find the final linear velocity, v = r*w_f.

Angular acceleration is directly given by a = α*r, where α is angular acceleration and r is the radius. Angular acceleration is also related to the linear (tangential) acceleration, a_t, by a = r*α, where α is angular acceleration and r is the radius. Tangential acceleration involves not only changes in speed (which causes tangential acceleration) but also changes in direction (which causes radial or centripetal acceleration).

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Answer : The rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]

Explanation :

As we are given the mechanism for the reaction :

Step 1 : [tex]2A\rightleftharpoons B+C[/tex]    (equilibrium)

Step 2 : [tex]B+D\rightarrow E[/tex]     (slow)

Overall reaction : [tex]2A+D\rightarrow C+E[/tex]

First we have to determine the equilibrium constant from step 1.

The expression for equilibrium constant will be,

[tex]K'=\frac{[B][C]}{[A]^2}[/tex]

Form this, the value of [B] is,

[tex][B]=\frac{K'[A]^2}{[C]}[/tex]       ............(1)

Now we have to determine the rate law from the slow step 2.

The expression for law will be,

[tex]Rate=K''[B][D][/tex]       .............(2)

Now put equation 1 in 2, we get:

[tex]Rate=K''\frac{K'[A]^2}{[C]}[D][/tex]

[tex]Rate=\frac{K[A]^2[D]}{[C]}[/tex]

Therefore, the rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]

The rate law for the reaction is obtained from the slow step (Step 2) and modified with information from the fast equilibrium step (Step 1), considering B as an intermediate reactant. The overall rate law becomes: Rate = k [ ((k1/k-1)[A]^2/[C]) ][D]

The rate law for a reaction is established based upon the rate-determining (slowest) step of the reaction. So, first, we identify the slow step of the reaction. From the information given, we can see that the slow step is Step 2, with reactants B and D forming product E.

The rate law for this step of the reaction is written as: Rate = k [B][D], where k is the rate constant for this reaction and [B] and [D] are the concentrations of reactants B and D respectively.

However, reactant B is an intermediate since it doesn’t appear in the overall reaction. So, the rate of formation of B in the fast equilibrium Step 1 becomes important. Here, B is produced from reactant A. The forward and backward rates are equal at equilibrium. So, the rate law for Step 1 could be written as: Rate = k1[A]^2 = k-1[B][C], giving us [B]= (k1/k-1)[A]^2/[C],

Substituting [B] in the rate law for step 2, we obtain the rate law for the overall reaction as: Rate = k [ ((k1/k-1)[A]^2/[C]) ][D], where k is the rate constant for the slow step and k1 and k-1 are the rate constants for the forward and reverse reactions of Step 1.

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Answer:

t=4.2 years

Explanation:

velocity v= 0.96c

destination star distance = 14.4 light year

According to the theory of relativity length contraction

[tex]l= \frac{l_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]l_0= l{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

putting values we get

[tex]l_0= 14.4{\sqrt{1-\frac{(0.96c)^2}{c^2}}}[/tex]

[tex]l_0= 4.032 light years[/tex]

now distance the trip covers

D= vt

[tex]4.032\times c= 0.96c\times t[/tex]

[tex]t= \frac{4.032c}{0.96c}[/tex]

t= 4.2 years

so the trip will take 4.2 years

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, [tex]a=49\ m/s^2[/tex]

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

[tex]F=75\ kg\times 49\ m/s^2[/tex]

F = 3675 N

Ratio, [tex]R=\dfrac{F}{W}[/tex]

[tex]R=\dfrac{3675}{75\times 9.8}=5[/tex]

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

[tex]F'=\sqrt{F^2+W^2}[/tex]

[tex]F'=\sqrt{(3675)^2+(75\times 9.8)^2}[/tex]

F' = 3747.7 N

The direction of force is calculated as :

[tex]\theta=tan^{-1}(\dfrac{W}{F})[/tex]

[tex]\theta=tan^{-1}(\dfrac{1}{5})[/tex]

[tex]\theta=11.3^{\circ}[/tex]

Hence, this is the required solution.

The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.

(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:

Force = mass * acceleration

Force = 75.0 kg * 49.0 m/s2

Force = 3675 N

Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:

Weight = mass * gravitational acceleration

Weight = 75.0 kg * 9.8 m/s2

Weight = 735 N

To find the ratio, we divide the force exerted by the seat by the weight of the passenger:

Ratio = Force / Weight

Ratio = 3675 N / 735 N

Ratio = 5

(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:

Total Force = √(horizontal component2 + vertical component2)

Total Force = √(36752 + 7352)

Total Force = 3793 N

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Answer:

The focal length of the eyepiece is 4.012 cm.

Explanation:

Given that,

Focal length = 1.5 cm

angular magnification = -58

Distance of image = 18 cm

We need to calculate the focal length of the eyepiece

Using formula of angular magnification

[tex]m=-(\dfrac{i-f_{e}}{f_{0}})\dfrac{N}{f_{e}}[/tex]

Where,

[tex]i[/tex] = distance between the lenses in a compound  microscope

[tex]f_{e}[/tex]=focal length of eyepiece

[tex]f_{0}[/tex]=focal length of the object

N = normal point

Put the value into the formula

[tex]-58=-(\dfrac{18-f_{e}\times25}{1.5\timesf_{e}})[/tex]

[tex]87f_{e}=450-25f_{e}[/tex]

[tex]f_{e}=\dfrac{450}{112}[/tex]

[tex]f_{e}=4.012\ cm[/tex]

Hence,  The focal length of the eyepiece is 4.012 cm.

Answer:

Torque = 99.48 N-m²

Explanation:

It is given that,

Radius of the flywheel, r = 1.93 m

Mass of the disk, m = 92.1 kg

Initial angular velocity, [tex]\omega_i=419\ rpm=43.87\ rad/s[/tex]

Final angular speed, [tex]\omega_f=0[/tex]

We need to find the constant torque required to stop it in 1.25 min, t = 1.25 minutes = 75 seconds

Torque is given by :

[tex]\tau=I\times \alpha[/tex]...........(1)

I is moment of inertia, for a solid disk, [tex]I=\dfrac{mr^2}{2}[/tex]

[tex]\alpha[/tex] is angular acceleration

[tex]I=\dfrac{92.1\ kg\times (1.93\ m)^2}{2}=171.53\ kgm^2[/tex]..............(2)

Now finding the value of angular acceleration as :

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

[tex]0=43.87+\alpha \times 75[/tex]

[tex]\alpha =-0.58\ m/s^2[/tex]..........(3)

Using equation (2) and (3), solve equation (1) as :

[tex]\tau=171.53\ kgm^2\times -0.58\ m/s^2[/tex]

[tex]\tau=-99.48\ N-m^2[/tex]

So, the torque require to stop the flywheel is 99.48 N-m². Hence, this is the required solution.

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

(a)The take-off speed is the speed at the start of takeoff. The take-off speed of the kangaroo will be 11 m/sec.

(b)The height achieved during takeoff is the maximum height. The maximum height above the ground will be 1.3 meters.

It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.The maximum height of motion is given by

[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]

The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula

[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]

(a)Take-of velocity =?

given

Horizontal range = 9.6m.

[tex]\theta = 28^0[/tex]

[tex]g = 9.81 \frac{m}{sec^{2} }[/tex]

[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]

[tex]u = \sqrt{\frac{Rg}{sin2\theta} }[/tex]

[tex]u = \sqrt{\frac{9.6\times9.81}{sin56^0} }[/tex]

[tex]u = 11 m /sec[/tex]

Hence the take-off speed of the kangaroo will be 11 m/sec.

(b) Maximum height =?

given,

[tex]u = 11 m /sec[/tex]

[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]

[tex]H = \frac{(11)^{2}sin^2 58^0 }{2\times 9.81}[/tex]

[tex]\rm { H = 1.3 meter }[/tex]

Hence the maximum height above the ground will be 1.3 meters.

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Answer: A) Quenching

Hope this helps

Answer :  The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of [tex]CsCl[/tex] :

(1) Conversion of solid calcium into gaseous cesium atoms.

[tex]Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)[/tex]

[tex]\Delta H_s[/tex] = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

[tex]Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)[/tex]

[tex]\Delta H_I[/tex] = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

[tex]Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)[/tex]

[tex]\Delta H_D[/tex] = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

[tex]Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)[/tex]

[tex]\Delta H_E[/tex] = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

[tex]Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)[/tex]

[tex]\Delta H_L[/tex] = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

[tex]\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L[/tex]

Now put all the given values in this equation, we get:

[tex]-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L[/tex]

[tex]\Delta H_L=-667KJ/mole[/tex]

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

The magnitude of  Lattice energy of CsCl is  -676 kJ/mol.

Given Here,

Enthalpy of sublimation of Cs  [tex]\rm \bold{ \Delta H(sub ) }[/tex] = +76 kJ/mo

Ionization Energy for Potassium  IE(Cs) = +376 kJ/mol

Electron affinity for Chlorine is EA(Cl)  =  −349 kJ/mol.

Bond dissociation energy of Chlorine, BE(Cl)  = +121 kJ/mol

Enthalpy of formation for CsCl,  [tex]\rm \bold{ \Delta H(f) }[/tex] =   −436.5 kj/mol .

The lattice energy of KCl can be calculated from the formula,

[tex]\rm \bold {U(CsCl) = \Delta Hf(CsCl) - [ \Delta H(sub) + IE(Cs) +BE(Cl_2) + EA(Cl)]}[/tex]

U( CsCl)  = -436 - [+76  +376   +121 kJ/mol  -349]

U( CsCl)  = -676 kJ/mol

Hence we can calculate that the magnitude of  Lattice energy of CsCl is  -676 kJ/mol.

To know more about Lattice Energy, refer to the link:

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Answer:

The current in wire 3 is 0.25 A.

Explanation:

It is given that, three wires meet at a junction.

Current in wire 1, I₁ = 0.4 A

Current in wire 2, I₂ = -0.65 A  (out of the junction)

We need to find the magnitude of the current in wire 3 (I₃). Applying Kirchhoff's current law which states that the sum of current in the circuit at a junction is equal to zero.

[tex]I_1+I_2+I_3=0[/tex]

[tex]I_3=-(I_2+I_1)[/tex]

[tex]I_3=-(-0.65+0.4)\ A[/tex]

I₃ = 0.25 A

So, the current in wire 3 is 0.25 A. Hence, this is the required solution.

Using the principle of Kirchhoff's junction rule, we calculate the current in Wire 3 as being 0.25 amperes.

The understanding of this question relies on the principle of Kirchhoff's junction rule in Physics, which states the sum of currents entering a junction equals the sum of currents leaving the junction. In this case scenario, Wire 1 has a current of 0.40 A into the junction and wire 2 has a current of 0.65 A out of the junction. The current of Wire 3 can be solved via the equation: I1 + I3 = I2, where I1, I2, and I3 are the currents on Wires 1, 2, and 3 respectively. Plugging in the given values, we get: 0.40 A + I3 = 0.65 A. Solving the equation for I3, we find that I3 = 0.25 A. Hence, the magnitude of the current in Wire 3 is 0.25 A.

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Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          [tex]\bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         [tex]\bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m[/tex]

The center of mass of the two-ball system is 7.05 m above ground.  

The resulting velocity of the center of mass is (A)[tex]11 m/s[/tex]downward.

To find the velocity of the center of mass of the two-ball system, we need to calculate the velocities of each ball after [tex]2.0[/tex] seconds and then use the center of mass formula.

[tex]v_1 = 0 + (9.8 \, \text{m/s}^2 \cdot 2.0 \, \text{s}) \\= 19.6 \, \text{m/s}[/tex](downward).

[tex]v_2 = 15 \, \text{m/s} - (9.8 \, \text{m/s}^2 \cdot 2.0 \, \text{s}) \\= 15 \, \text{m/s} - 19.6 \, \text{m/s} \\= -4.6 \, \text{m/s}[/tex]

(negative since it's moving downward).

Total mass ([tex]M[/tex]): [tex]0.50 kg + 0.25 kg = 0.75 kg[/tex]

[tex]V_{\text{cm}} = \frac{0.50 \, \text{kg} \cdot 19.6 \, \text{m/s} + 0.25 \, \text{kg} \cdot (-4.6 \, \text{m/s})}{0.75 \, \text{kg}} \\= \frac{9.8 - 1.15}{0.75} \\= 11.0 \, \text{m/s}[/tex] (downward)

Complete question:-

At the same instant that a [tex]0.50kg[/tex] ball is dropped from [tex]25 m[/tex] above Earth, a second ball, with a mass of [tex]0.25 kg[/tex], is thrown straight upward from Earth's surface with an initial speed of [tex]15 m/s.\\[/tex]. They move along nearby lines and pass without colliding. At the end of [tex]2.0 s[/tex] the velocity of the center of mass of the two-ball system is:

A) [tex]11 m/s[/tex], down

B) [tex]11 m/s[/tex], up

C) [tex]15 m/s[/tex], down

D) [tex]15 m/s[/tex], up

E) [tex]20 m/s[/tex], down

Final answer:

The angle a vector makes with the positive x-axis can be found using the cosine ratio; the angle is approximately 58.2 degrees for the vector with a magnitude of 25 m and an x-component of 12 m.

Explanation:

To determine the angle a vector makes with the positive x-axis, we can use trigonometry since we have the magnitude of the vector and the length of the x-component. Specifically, the cosine of the angle (θ) that the vector makes with the x-axis is equal to the x-component divided by the magnitude of the vector.

Here is the formula:

cos(θ) = (x-component) / (magnitude)

By substituting the given values:

cos(θ) = 12 m / 25 m

Now calculate the angle:

θ = cos⁻¹(12/25)

θ = 58.2° (approx.)

Therefore, the vector makes an angle of approximately 58.2 degrees with the positive x-axis.