A block weighing 87.0 N rests on a plane inclined at 24.1° to the horizontal. The coefficient of the static and kinetic frictions are 0.25 and 0.13 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?

Answer:

Maximum separation between buildings, D = 1.806 m

Explanation:

Vertical motion of criminal

Initial speed = 0 m/s

Acceleration = 9.81 m/s²

Displacement = 2.1 m

We need to find time when he moves vertically 2.1 m.

We have

         s = ut + 0.5at²

        2.1 = 0 x t + 0.5 x 9.81 x t²

           t = 0.43 s.

Horizontal motion of criminal

Initial speed = 4.2 m/s

Acceleration =0 m/s²

Time = 0.43 s

We need to find displacement.

We have

         s = ut + 0.5at²

        s = 4.2 x 0.43 + 0.5 x 0 x 0.43²

       s = 1.806 m

So maximum separation between buildings, D = 1.806 m

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:

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

Final answer:

The Physics question involves calculating the total momentum before and after a traffic collision. Total momentum before the collision is 76800 kg·m/s northward. Assuming a perfectly inelastic collision where the vehicles stick together, the total momentum remains unchanged after the collision.

Explanation:

To calculate the total momentum before the collision, we use the formula: momentum = mass × velocity. For the 1200 kg car traveling north at 14 m/s, its momentum is 1200 kg × 14 m/s = 16800 kg·m/s northward.

For the 2000 kg truck traveling at 30 m/s, its momentum is 2000 kg × 30 m/s = 60000 kg·m/s northward. The total momentum before the collision is the sum of these two momenta, so 76800 kg·m/s northward.

Assuming the collision is perfectly inelastic and the two vehicles stick together, the law of conservation of momentum states that the total momentum before the collision must be equal to the total momentum after the collision.

Therefore, the total momentum after the collision will also be 76800 kg·m/s northward.

Answer:

Distance covered by the car is 56.01 feet.

Explanation:

It is given that,

Initial velocity, u = 50 mi/h = 73.33 ft/s

Constant deceleration of the car, [tex]a=-48\ ft/s^2[/tex]

Final velocity, v = 0

Let s is the distance covered before the car comes to rest. It can be calculated using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0^2-(73.33\ ft/s^2)^2}{2\times -48\ ft/s^2}[/tex]

s = 56.01 ft

So, the distance covered before the car comes to a stop is 56.01 feet. Hence, this is the required solution.

The distance covered by the car as it comes to rest is 17.1 m.

To calculate the distance covered before the car come to stop, we use the formula below.

Formula:

Where:

make s the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, the distance covered by the car as it comes to rest is 17.1 m.

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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:

The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]

Explanation:

Given that,

Upward current = 24 A

Force per unit length[tex]\dfrac{F}{l} =88\times10^{4}\ N/m [/tex]

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force

[tex]F=ILB[/tex]

[tex]\dfrac{F}{l}=\dfrac{\mu I_{1}I_{2}}{2\pi r}[/tex]

Where,

[tex]\dfrac{F}{l}[/tex]=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

[tex]88\times10^{4}=\dfrac{4\pi\times10^{-7}\times24\times I_{2}}{2\pi \times7\times10^{-2}}[/tex]

[tex]I_{2}=\dfrac{88\times10^{4}\times7\times10^{-2}}{2\times\times10^{-7}\times24}[/tex]

[tex]I_{2}=1.3\times10^{10}\ A[/tex]

Hence, The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]

Answer:

option (B) - true

option (C) - true

option (D) - true

option (E) - true

option (F) - true

Explanation:

Answer:

2.3 kg

Explanation:

L = length of the plank = 4 m

M = mass of the plank = 7 kg

m = mass of cat = ?

[tex]F_{c}[/tex] = Weight of the cat = mg

[tex]F_{p}[/tex] = Weight of the plank = Mg = 7 x 9.8 = 68.6 N

ED = 2 m

CD = 1.5 m

EC = ED - CD = 2 - 1.5 = 0.5 m

Using equilibrium of torque about sawhorse at C

[tex]F_{p}[/tex] (EC) = [tex]F_{c}[/tex] (CD)

(68.6) (0.5) = (mg) (1.5)

(68.6) (0.5) = (m) (1.5) (9.8)

m = 2.3 kg

Final answer:

To determine the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper, use the formula L = (R·A)/ρ. The length of the copper wire, based on the provided resistance of 0.172 Ω, cross-sectional area of 7.85 × [tex]10^-^5[/tex] m², and resistivity of 1.72 × [tex]10^-^8[/tex]Ω·m, is calculated to be 7.85 meters.

Explanation:

The question involves determining the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper. The formula to calculate the length of the wire is derived from Ohm's law, which in terms of resistivity (ρ) is expressed as R = ρL/A, where R is the resistance, L is the length of the wire, A is the cross-sectional area, and ρ is the resistivity of the material.

We are provided with the resistance (R) as 0.172 Ω, the cross-sectional area (A) as 7.85 × [tex]10^-^5[/tex] m², and the resistivity (ρ) of copper as 1.72 × [tex]10^-^8[/tex] Ω·m. By rearranging the formula to solve for L, we get L = (R·A)/ρ. Substituting the given values into the formula, we find the length of the copper wire.

Let's calculate: L = (0.172 Ω × 7.85 × [tex]10^-^5[/tex] m^2) / (1.72 × [tex]10^-^8[/tex] Ω·m) = 7.85 m. Therefore, the length of the copper wire is 7.85 meters.

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:

The velocity of the boat after Batman lands in it is 2.08 m/s.

Explanation:

Given that,

Mass of batman [tex]m_{1}= 88.8\ kg[/tex]

Mass of boat [tex]m_{2}=530\ kg[/tex]

Initial velocity = 14.5 m/s

We need to calculate the velocity of boat

Using conservation of momentum

[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)

Where, [tex]m_{1}[/tex]=mass of batman

[tex]m_{2}[/tex] =mass of boat

u=initial velocity

v = velocity of boat

Put the value in the equation

[tex]88.8\times14.5=530+88.8\times v[/tex]

[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]

[tex]v=2.08\ m/s[/tex]

Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.

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:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

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.

Magnitude of force on each chain suspended with 100 ib load in a room, must be support is 314.54 N.

If the load, is hanging on chain or rope, and is in the equilibrium state, then the sum of all the horizontal component of it must is equal to zero.

Given information-

The load suspended by two chains in a room is 100-ib.

The angle between each chain and the horizontal ceiling is 45°.

The image attached below shows the given situation.

First convert the 100-ib into unit of N as,

[tex]F=100\rm ib=100\times4.4482\rm N\\F=444.82N[/tex]

As the load suspended by two chains in a room is 100-ib and the angle between each chain and the horizontal ceiling is 45°. Thus, the vertical component of the,

[tex]\sin (45)=\dfrac{F_y}{444.82}\\F_y=314.54\rm N[/tex]

Hence, the magnitude of the force each chain must be support is 314.54 N.

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Thus, the magnitude of the force each chain must support is approximately 70.7 lb.

To find the magnitude of the force each chain must support, you can use the principles of equilibrium.

Given:

Since the system is in equilibrium, the vertical forces must balance the weight of the load, and the horizontal components of the forces must cancel each other out.

1. Determine the vertical component of the force in each chain:

Each chain supports part of the total load. Let [tex]\( F \)[/tex] be the force in each chain.

Since the chains are symmetric, each chain supports half the load.

The vertical component of the force in each chain is given by:

[tex]\[ F_{\text{vertical}} = F \sin(45^\circ) \][/tex]

where [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex].

2. Calculate the vertical component for each chain:

For equilibrium, the sum of the vertical components of the forces in the two chains must equal the total load:

[tex]\[ 2 \times F \sin(45^\circ) = 100 \text{ lb} \][/tex]

Substitute [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex] :

[tex]\[ 2 \times F \times \frac{\sqrt{2}}{2} = 100 \][/tex]

Simplify:

[tex]\[ F \sqrt{2} = 100 \][/tex]

Solve for [tex]\( F \)[/tex]:

[tex]\[ F = \frac{100}{\sqrt{2}} \][/tex]

Rationalize the denominator:

[tex]\[ F = \frac{100 \sqrt{2}}{2} = 50 \sqrt{2} \][/tex]

3. Approximate the force:

Using [tex]\( \sqrt{2} \approx 1.414 \)[/tex]:

[tex]\[ F \approx 50 \times 1.414 = 70.7 \text{ lb} \][/tex]

The ball thrown by a major-league pitcher to a catcher 17.0m away, with a speed of 41.0m/s, would drop approximately 0.84 meters due to the gravitational acceleration.

The subject of this question involves physics, specifically the concept of projectile motion. When a pitcher throws a ball with a horizontal speed, the ball also experiences a vertical motion due to gravity. Therefore, even though the ball is thrown horizontally, it will gradually drop as it travels towards the catcher. To figure out how far it drops, we can use the physics equation for motion under constant acceleration: distance = 0.5 * acceleration * time2.

Here, the acceleration is due to gravity, which is approximately 9.8m/s2. The time can be calculated by dividing the horizontal distance (17.0 m) by the horizontal speed (41.0 m/s), giving us approximately 0.4146 s. Substituting these into the equation: 0.5 * 9.8m/s2 * (0.4146s)2 = 0.84 meters, the drop of the ball is approximately 0.84 meters.

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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:

hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]

Explanation:

shift in wavelength due to compton effect is given by

[tex]\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )[/tex]

λ' = the wavelength after scattering

λ= initial wave length

h= planks constant

m_{e}= electron rest mass

c= speed of light

θ= scattering angle = 180°

compton wavelength is

[tex]\frac{h}{m_{e}c}= 2.43\times10^{-12}m[/tex]

[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )[/tex]

[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )[/tex]  ( put cos 180°=-1)

also given λ'=2λ

putting values and solving we get

[tex]\lambda =4.86\times10^{-12}m[/tex]

hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]

Answer:

≈1,39 and 4.56 kg.

Explanation:

for more details see the attachment. Note, the ρ(Cu) means 'Density of copper', ρ(Al) means 'Density of aluminium'.

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

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:

Force, [tex]F=8.23\times 10^{-8}\ N[/tex]

Explanation:

It is given that,

Each ion in Na⁺ and Cl⁻ has a charge of, [tex]q=1.6\times 10^{-19}\ C[/tex]

Distance between two ions, [tex]d=5.29\times 10^{-11}\ m[/tex]

We need to find the electrostatic force. It is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{(5.29\times 10^{-11})^2}[/tex]

[tex]F=8.23\times 10^{-8}\ N[/tex]

So, the magnitude of electrostatic force between them is [tex]F=8.23\times 10^{-8}\ N[/tex]. Hence, this is the required solution.

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: a) 7,00 centimeters

(b) 259. 19 feet

(c) 3110.28 inches

(d) 0.049 miles

Explanation:

(a) We know that 1 meter = 100 centimeters

Therefore,

[tex]79\ m= 7,900\text{ centimeters}[/tex]

(b)Since 1 meter = 3.28084 feet

Then, [tex]79\ m= 79\times3.28084=259.18636\approx 259.19\text{ feet}[/tex]

(c) Since, 1 feet = 12 inches.

[tex]79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}[/tex]

(d) [tex]\text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}[/tex]

[tex]79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}[/tex]

Answer:

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Explanation:

(a) The velocity of the airplane relative to the skydiver is 52 m/s, and the acceleration is directed radially inward. (b) The time rate of change of the speed [tex]\(v_r\)[/tex] of the airplane, as observed by the skydiver, is 0 m/s², and the radius of curvature [tex]\(\rho_r\)[/tex] of its path is 2330 m.

(a) The velocity of the airplane relative to the skydiver is the vector difference of their individual velocities. Since both have the same constant speed of 52 m/s, the relative velocity is 52 m/s. The acceleration of the airplane, as observed by the skydiver, is directed radially inward due to the circular motion.

(b) The time rate of change of speed [tex](\(v_r\))[/tex] is the radial component of acceleration, which is 0 m/s² since the airplane is moving at a constant speed. The radius of curvature [tex](\(\rho_r\))[/tex] of its path is given as 2330 m, representing the circular path's curvature.

These results are derived from principles of relative motion and circular motion, providing insights into the dynamics of the airplane-skydiver system.

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.

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Answer:

[tex]Q = 8.61 \times 10^{-4} C[/tex]

Explanation:

Since in LC oscillation there is no energy loss

so here we can say that

initial total energy of capacitor = energy stored in capacitor + energy stored in inductor at any instant of time

so we can say

[tex]\frac{Q^2}{2C} = \frac{q^2}{2C} + \frac{1}{2}Li^2[/tex]

now we have

[tex]q = 3\mu C[/tex]

[tex]i = 75 \mu A[/tex]

now we have

[tex]Q^2 = q^2 + (LC) i^2[/tex]

we also know that

[tex]2\pi f = \frac{1}{\sqrt{LC}}[/tex]

[tex]2\pi(1.6) = \frac{1}{\sqrt{LC}}[/tex]

[tex]LC = 9.89 \times 10^{-3}[/tex]

now from above equation

[tex]Q^2 = (3\mu C)^2 + (9.89 \times 10^{-3})(75 \mu A)[/tex]

[tex]Q = 8.61 \times 10^{-4} C[/tex]

To find the maximum charge of the capacitor in the LC circuit, set up a cosine equation using the initial charge and current value.

To find the maximum charge of the capacitor in the LC circuit, we need to utilize the relationship between the charge on the capacitor and the current in the circuit. At time t = 0, the capacitor is fully charged, so the initial charge is Q = 3 µC. The current in the circuit is equal to 75 µA. Knowing these values, we can set up a cosine equation to find q(t). By solving the equation, we can find the maximum charge of the capacitor.

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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.

To learn more about the range of projectile refer to the link ;

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