An electromagnet produces a magnetic field of 0.520 T in a cylindrical region of radius 2.40 cm between its poles. A straight wire carrying a current of 10.5 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field.
What magnitude of force is exerted on the wire?

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

[tex]ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m[/tex]

The spring's force constant is 13.54 N/m

Total energy of the system is given by

[tex]E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J[/tex]

At maximum displacement v = 0

[tex]E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)[/tex]

[tex]\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m[/tex]

The amplitude of the motion is 0.6 m

Speed of the block

[tex]E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s[/tex]

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

[tex]ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2[/tex]

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²

The force constant of the spring is 13.54 N/m. The amplitude of the oscillatory motion is 0.240 m. The maximum velocity and acceleration are 4.72 m/s and 13 m/s² respectively.

First, we can calculate the spring's force constant k using Newton's second law (F = ma) and Hooke's Law (F = kx), by equating the forces F (Note that the negative acceleration is due to the force acting opposite to displacement). Hence, having the force due to spring F = m * a = 0.250 kg * -13.0 m/s² = -3.25 N and applying force due to spring as per Hooke's law as -k * x. Equating these forces gives k = 13.54 N/m.

Secondly, the amplitude A of the motion can be identified as |x| = 0.240 m, since for a simple harmonic oscillator, the amplitude is the maximum displacement from equilibrium (x = 0).

For the third part, the maximum speed Vmax during its motion can be calculated using the energy conservation principle, as given by the formula, Vmax = sqrt(k/m * A²), which gives Vmax = 4.72 m/s.

Finally, the maximum acceleration of the block Amax happens at the points of maximum displacement, i.e., at the amplitude points, which is Amax = k/m * A = 13m/s².

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Answer:Hollow sphere

Explanation:

Given

same mass for solid and hollow sphere

same [tex]v_{cm}[/tex] before they start up incline

Moment of inertia of solid Sphere

[tex]I_1=\frac{2}{5}Mr^2[/tex]

Moment of inertia of hollow sphere

[tex]I_2=\frac{2}{3}Mr^2[/tex]

Conserving Energy at bottom and top point for solid sphere

kinetic energy +Rotational Energy=Potential energy

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1[/tex]

for pure rolling [tex]v_{cm}=\omega r[/tex]

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1[/tex]

[tex]\frac{7}{10}Mv_{cm}^2=Mgh_1[/tex]

[tex]h_1=\frac{7v_{cm}^2}{10g}[/tex]

For hollow sphere

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2[/tex]

[tex]h_2=\frac{5v_{cm}^2}{6g}[/tex]

therefore height gained by hollow sphere is more

The hollow sphere reached the greatest height.

The height reached by each object is determined by applying the principle of conservation of energy as show below;

K.E = P.E

¹/₂mv² + ¹/₂Iω² = mgh

where;

I =  ²/₅mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh

¹/₂v² + ¹/₅v² = gh

7v² = 10gh

h = 7v²/10g

h = 0.7(v²/g)

I = ²/₃mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh

¹/₂v² + ¹/₃v² = gh

5v² = 6gh

h = 5v²/6g

h = 0.83(v²/g)

Thus, the hollow sphere reached the greatest height.

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Answer:

a) 24 Hs. b) 0.224 m/s² c) 448 N

Explanation:

a) As satellites in a geosynchronous orbits, stay directly over a point fixed on the Equator while the Earth rotates, the only way that this can be possible, if the period of the satellite (time to complete a full orbit) is equal to the time that the Earth uses to complete a spin itself, which is exactly one day.

b)

The value of g, is just the acceleration due to the gravitational attraction between the satellite and the Earth.

According the Universal Law of Gravitation, this force can be written in this way:

Fg = ms . a = G me. ms / (re+rs)² ⇒a=g= G me / (re + rs)²

Replacing by the values of G, me, re, and rs, we get:

g = 6.67. 10⁻¹¹ . 5.97.10²⁴ / (6.37 10⁶ + 3.58.10⁷)² m/s²

g= 0.224 m/s²

c) If we call "weight" to the magnitude of the gravitational force on the satellite (as we do with masses on Earth), we can find this value, just solving the equation for Fg, as follows:

Fg = G me . ms / (re + rs)²

Replacing by the values, we find:

Fg = 448 N

A satellite in geosynchronous orbit has a period of 24 hours. The gravitational acceleration at this altitude is roughly 0.224 m/s^2. The weight of a 2000 kg satellite in a geosynchronous orbit is roughly 448 N.

A satellite in a geosynchronous orbit has a period of 24 hours, which equates to how long it takes the Earth to rotate once on its axis. The value of gravitational acceleration g at an altitude of a geosynchronous orbit can be calculated using the formula g = G × (M / (R+h)^2), where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth, and h is the altitude of the satellite.

Using the given altitude, the value of g at this height is approximately 0.224 m/s^2. Now, to find the weight of a 2000 kg satellite in a geosynchronous orbit, we use the formula W=mg where m is the mass of the satellite and g is the gravitational acceleration at the altitude. That gives us the value of roughly 448 N for the weight of the satellite at geosynchronous orbit.

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The smallest distance from the wall where a listener could hear constructive interference from two in-phase loudspeakers emitting sound at a frequency of 514 Hz would be 0.668 m, under the conditions given in the problem.

The phenomenon being described here is constructive interference, which occurs when two waves overlap so that their combined effect is stronger than their individual effects. In this scenario, we calculate the path difference, which is the difference in the path traveled by the sound from the two loudspeakers. Constructive interference occurs when the path difference corresponds to an integer number of wavelengths.

First, let's calculate the wavelength of the sound: λ = speed of sound / frequency. So, λ = 343 m/s / 514 Hz = 0.668 m.

For constructive interference, the path difference should be Nλ, where N is an integer. Since we're looking for the smallest possible distance, we should take N = 1. So, the distance from the wall where the listener stands (d = path difference) should be 1 × λ = 1 × 0.668 m = 0.668 m. But none of the options given match this value. Therefore, none of the options given are correct, if we correctly followed the conditions given in the problem.

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The person standing in front of the loudspeakers, which are emitting sound waves with a frequency of 514 Hz, will experience constructive interference at a distance closest to 1.15 m from one of the speakers.

In this case, the person standing in front of the loudspeakers will experience constructive interference - where the waves from the two loudspeakers are in-phase and add up to make a louder sound - at distances that are multiples of one wavelength away from one speaker. The wavelength (λ) of the sound can be calculated using the formula λ = v / f, where v is the speed of sound and f is the frequency. Substituting the given values, λ = 343 m/s / 514 Hz = 0.667 m.

The student will hear constructive interference at distances of nλ/2 from the speaker, where n is an integer. The smallest non-zero value for n is 1, so the closest such point will be 0.667 m / 2 = 0.334 m. However, this is less than the smallest distance in the multiple choice answers, which suggest the answer should be in metres. The next value for n is 2, which gives a distance of 0.667 m, still less than the smallest multiple choice answer. The next value for n is 3, which gives a distance of 1.001 m, which falls between the 1st and 2nd options so the answer should correspond to the closest choice which is 1.15 m.

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Answer:

$364.29

Explanation:

given,

Packing of crates per month (u)= 800

annual carrying cost of 35 percent of the purchase price per crate.

Ordering cost(S) = $ 28

D = 800 x 12 = 9600 crates/year

H = 0.35 P

H = 0.35 x $10

H = $3.50/crate per yr.

Present Total cost

= [tex]\dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28[/tex]

= 1400 + 336

= $ 1,736

[tex]Q_0 = \sqrt{\dfrac{2DS}{H}}[/tex]

[tex]Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}[/tex]

[tex]Q_0 =\$ 391.92[/tex]

Total cost at EOQ

= [tex]\dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28[/tex]

= 685.86 + 685.85

= $ 1,371.71

the firm save annually in ordering and carrying costs by using the EOQ

    = $ 1,736 - $ 1,371.71

    = $364.29

The question focuses on Economic Order Quantity (EOQ) in inventory management for a produce distributor's packing crates. EOQ helps determine the optimal number of units a company should add to its inventory to minimize total holding and setup costs.

The subject of this question is

Economic Order Quantity (EOQ)

, which is an operational efficiency measure used in inventory management. It refers to the number of units that a company should add to its inventory at a time to minimize total holding and setup costs.

In this case, the produce distributor uses 800 crates a month, purchased at $10 each. The annual carrying cost is 35% of the purchase price per crate, and the ordering cost is $28. Currently, the distributor orders once a month. Hence, to determine the optimal number of orders, or the EOQ, various factors such as ordering cost, holding cost, and demand rate need to be considered.

This cost minimization issue can be mathematically represented and solved using the EOQ formula: EOQ = √((2DS)/H) where D is the demand rate, S is the setup (or order) cost, and H is the holding (or carrying) cost.

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Answer:

439.587 kgm²

Explanation:

[tex]m_p[/tex] = Mass of person = 71 kg

R = Radius of platform = 1.62 m

[tex]m_d[/tex] = Mass of disc = 193 kg

The moment of inertia of the system is given by

[tex]I=m_pR^2+\frac{1}{2}m_dR^2\\\Rightarrow I=71\times 1.62^2+\frac{1}{2}\times 193\times 1.62^2\\\Rightarrow I=439.587\ kgm^2[/tex]

The total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk is 439.587 kgm²

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

(A) The force constant of the spring is obtained as 34.66 N/m

(B) The unloaded length of the spring is 20 cm.

(A) According to Hooke's law;

[tex]F = kx[/tex]

Where 'F' is the force needed to extend or compress the spring, 'k' is the force constant and 'x' is the extended length or compressed length.

Here, the gravitational force of the mass provides the necessary force for the spring to elongate.

Therefore, we can say that in the case of first mass;

[tex]k\,(0.25-l_0) = m_1g = (0.175\,kg \times 9.8\,m/s^2)[/tex]

Where [tex]l_0[/tex] is the unloaded length of the spring.

[tex]\implies k\,(0.25-l_0)=1.715\,N[/tex]

In the case of the second mass, we can write;

[tex]k\,(0.775\,m-l_0) = m_2g = (2.075\,kg \times 9.8\,m/s^2)[/tex]

[tex]\implies k\,(0.775\,m-l_0) = 20.335\,N[/tex]

From both these equations, we can write;

[tex]k\,(0.775\,m-l_0-0.25\,m+l_0)=20.335\,N- 1.715\,N[/tex]

[tex]\implies0.525\, k=18.2\,N[/tex]

[tex]\therefore k=\frac{18.2\,N}{0.525\,m}=34.66\,N/m[/tex]

(B) Applying the value of 'k' in any of the equations for force of given masses, we get;

[tex]k\,(0.25-l_0)=1.715\,N[/tex]

[tex]\implies (34.66\,N/m)\times\,(0.25-l_0)=1.715\,N[/tex]

[tex]\implies 8.665N-34.66\,l_0=1.715\,N[/tex]

[tex]\implies \,l_0=\frac{-6.95N}{-34.66} =0.2\,m=20\,cm[/tex]

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Answer:

D. There is no way to know which skier has the greater speed at the finish.

Explanation::

Total energy is conserved because there is no friction

(Et)₀ = (Et)f

(Et)₀ = Initial total energy (J)

(Et)f  = Final total energy (J)

(Et)₀= K₀ + U₀

(Et)f= Kf + Uf

K₀ : Initial kinetic energy

U₀ : Initial potential energy

Kf  : final kinetic energy

Uf :  final potential energy

The formulas to calculate the kinetic energy (K) and potential energy (U) are:

K = ( 1/2)m*v²

U = m* g*h

m : mass (kg)

v: speed ( m/s)

h: hight ( m)

Problem development

Skier A

(Et)₀ = (Et)f

K₀ + U₀ =  Kf + Uf

( 1/2)mA*(v₀A)² + mA*g*h₀ =  ( 1/2)mA*(vfA)² + mA*g*hf ,

We divide by mA on both sides of the equation

( 1/2)*(v₀A)²+ g*h₀ =  ( 1/2)(vfA)² + g*hf

( 1/2)*(v₀A)²+ g*h₀ - g*hf  =  ( 1/2)(vfA)²  

We multiply by 2 both sides of the equation

(v₀A)²+2g(h₀ -hf) = (vfA)²  

(vfA)²  = (v₀A)²+2g(h₀ -hf) Equation (1)

Skier B

(Et)₀ = (Et)f

K₀ + U₀ =  Kf + Uf

(1/2)mB*(v₀B)² + mB*g*h₀ =  ( 1/2)mB*(vfB)² + mB*g*hf

We perform the same procedure above:

(vfB)²  = (v₀B)²+2g(h₀ -hf) Equation (2)

Comparison of equation (1) with equation (2)

The term 2g (h₀ -hf) is the same in both equations because the paths of the two skiers start in the same place and end in the same place.

The final speed (vf) of skiers depends on their initial speed (v₀).

Because the initial speed of the skiers is unknown, it cannot be determined which has the highest final speed

To solve this problem it is necessary to apply the related concepts to the scalar value of displacement current, which can be expressed in terms of electric flux as

[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]

Where,

[tex]\epsilon_0[/tex] = Permitibitty of free space constant

[tex]\Phi_E[/tex] = Magnetic flux

t = time

We know as well that the Flux can be expressed as

[tex]\Phi = EA[/tex]

Here

A= Cross-sectional area

E = Electric Potential in a Uniform electric field

At the same time the electric potential is expressed in terms of Voltage and distance, that is

[tex]E = \frac{V}{d}[/tex]

Using this equation we have then that

[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]

[tex]I_d = \epsilon_0 \frac{d(EA)}{dt}[/tex]

[tex]I_d = \epsilon_0*A (\frac{d(E)}{dt})[/tex]

[tex]I_d = \epsilon_0*A (\frac{d(V)}{dt*d})[/tex]

[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]

According to our values we have that

[tex]\frac{dV}{dt} = 107V/s[/tex]

[tex]A = 0.174m^2[/tex]

[tex]d = 1.06*10^{-3}m[/tex]

[tex]\epsilon = 8.85418^{-12} m^{-3}kg^{-1}s^4A^2[/tex]

Replacing,

[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]

[tex]I_d = \frac{(8.85418^{-12})*(0.174)}{1.06*10^{-3}} (107)[/tex]

[tex]I_d = 7.565*10^{-8}A[/tex]

Therefore the displacement current is [tex]7.565*10^{-5}mA[/tex]

The displacement current (in mA) between these plates is equal to [tex]1.56 \times 10^{-5}\;mA[/tex]

Given the following data:

Scientific data:

Mathematically, the displacement current (in mA) between the plates in an electric field is given by this formula:

[tex]I_d=\frac{\epsilon _o A}{d} (\frac{d(v)}{dt} )\\\\[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]I_d=\frac{8.854 \times 10^{-12} \times 0.174 \times 107}{1.06 \times 10^{-2}} \\\\I_d=1.56 \times 10^{-8}\\\\I_d=1.56 \times 10^{-5}\;mA[/tex]

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Answer

given,

diameter of the steel ball = 3 m = 0.003 m

density of the steel = 7600 kg/m³

density of liquid = 1200 kg/m³

distance travel by the ball = 0.5 m

time = t = 10 s

average velocity =[tex]v = \dfrac{0.5}{10}[/tex]

  v = 0.05 m/s

a) density of water is less than ball so, ball will fall in the fluid.

gravitational force is equal to buoyancy force and the drag force

[tex]F_g = F_b + F_d[/tex]

[tex]F_g = \rho_g Vg[/tex]

Density of ball = ρ_s

V is the volume ball

buoyancy force

[tex]F_b = \rho_f Vg[/tex]

[tex]F_b = \rho_f V g[/tex]

drag force

 [tex]F_d =3 \pi \mu d v[/tex]

[tex]F_g = F_b + F_d[/tex]

[tex] \rho_g Vg = \rho_f V g+ 3 \pi \mu d v[/tex]

[tex] (\rho_g - \rho_f)Vg = 3 \pi \mu d v[/tex]

[tex]V= \dfrac{1}{6}\pi d^3[/tex]

[tex] (\rho_g - \rho_f). \dfrac{1}{6}\pi d^3.g = 3 \pi \mu d v[/tex]

[tex]\mu = \dfrac{(\rho_s-\rho_f)d^2g}{18 v}[/tex]

[tex]\mu = \dfrac{(7600 -1200)\times 0.003^2\times 9.8 }{18 \times 0.05}[/tex]

μ = 0.63 kg m/s

Answer:

X rays, ultraviolet light, green light, red light, infrared radio waves

Explanation:

The electromagnetic spectrum can be thought of as being arranged either in decreasing frequency or increasing wavelength. The full spectrum is as follows, from the shortest wavelength to the longest:

Gamma rays, X rays, ultraviolet light, violet light, indigo light, blue light, green light, yellow light, orange light, red light, infrared,  radio waves, radar waves, microwaves, television waves, radio waves.

The electromagnetic radiation types can be arranged from shortest to longest wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, and radio waves.

The electromagnetic radiation types provided can be arranged in order of increasing wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, radio waves. This order represents the electromagnetic spectrum from shortest to longest wavelengths. In simpler terms, X rays have the shortest wavelength, followed by ultraviolet light, green light, red light, infrared, and finally radio waves with the longest wavelength. Wavelength and frequency are inversely related - the shorter the wavelength, the higher the frequency.

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Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 32}[/tex]

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

[tex]J=m(v-u)[/tex]

[tex]J=1.2\times (-25.04-10)[/tex]

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

[tex]J=\dfrac{F}{t}[/tex]

[tex]F=J\times t[/tex]

[tex]F=42.048\times 0.02[/tex]

F = 0.8409 N

Hence, this is the required solution.

Answer:

7.72 Liters

Explanation:

normal body temperature = T_body =37° C

temperature of ice water = T_ice =0°c

specfic heat of water = c_{water} =4186J/kg.°C

if the person drink 1 liter of cold water mass of water is = m = 1.0kg

heat lost by body is Qwater =mc_{water} ΔT

                                           = mc{water} ( T_ice - T_body)

                                             = 1.0×4186× (0 -37)

                                             = -154.882 ×10^3 J

here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories

            = 286×4186J

so number of liters of ice water have to drink is

n×Q_{water} =Q_{walk}                                                                        n= Q_{walk}/ Q_{water}

= 286×4186J/154.882×10^3 J

 = 7.72 Liters

Answer:

[tex]V=7.73\ L[/tex]

Explanation:

Given:

Initial temperature of water,  [tex]T_i=0^{\circ}C[/tex]

final temperature of water, [tex]T_f=37^{\circ}C[/tex]

energy spent in one hour of walk, [tex]286\ kilocal=(286\times 4186)\ J[/tex]

volumetric capacity of stomach, [tex]V=1\ L[/tex]

Now, let m be the mass of water at zero degree Celsius to be drank to spend 286 kilo-calories of energy.

[tex]\therefore Q=m.c_w.\Delta T[/tex] .....................................(1)

where:

m = mass of water

Q = heat energy

[tex]c_w=4186\ J\ (specific\ heat\ of\ water)[/tex]

[tex]\Delta T[/tex]= temperature difference

Putting values in the eq. (1):

[tex]286\times 4186=m\times 4186\times 37[/tex]

[tex]m=7.73\ kg[/tex]

Since water has a density of 1 kilogram per liter, therefore the volume of water will be:

[tex]V=7.73\ L[/tex]

Answer:

option A

Explanation:

given,

velocity to hit home plate = 110 Km/h

a) When the baseball is hit straight back

     Assuming the momentum before hitting be P and after hitting the bal is also equal to P.

change in momentum = P - (-P) = 2P

b) When the catcher catches the ball the change in momentum is equal to zero.

c) when the ball is popped up the change in momentum

   p_x = P and P_y = P

  resultant momentum = [tex]\sqrt{P^2+ P^2} = \sqrt{2}\ P[/tex]

the maximum change in momentum will be in case  A

The correct answer is option A

Final answer:

The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h.

Explanation:

The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h. This is because when the baseball is hit straight back, it completely reverses its direction of motion, resulting in a large change in momentum. The other scenarios, such as the baseball being caught by the catcher or popped straight up, do not involve a complete reversal of direction, so the change in momentum is smaller.

To solve this problem it is necessary to apply the concepts of thermal expansion. Thermal expansion can be expressed in longitudinal terms such as

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

Where,

[tex]\alpha =[/tex] Thermal Expanssion Coefficient

[tex]L_0 =[/tex] Initial Length

[tex]\Delta T =[/tex] Change in Temperature

Our values are given as

[tex]\alpha = 11*10^{-6}/\°C \rightarrow[/tex] from Steel

[tex]L_0 = 15m[/tex]

[tex]T_1 = -23\°C[/tex]

[tex]T_2 = 32\°C[/tex]

Replacing we have that,

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

[tex]\Delta L = (17)(11*10^{-6})(32-(-21))[/tex]

[tex]\Delta L = 0.009911m[/tex]

[tex]\Delta L = 9.911mm[/tex]

Therefore the difference in length of the beam between these two temperature extremes is 9.911mm

To solve this problem it is necessary to apply the concepts related to heat transfer and power depending on energy and time.

By definition we know that the heat loss of water is given by

[tex]Q = mc_w*\Delta T+m*L_f[/tex]

Where,

m = mass

[tex]c_w =[/tex] Specific Heat of Water

T = Temperature

[tex]L_f =[/tex]Latent heat of fusion [tex] \rightarrow[/tex] Heat of fusion for water at 0°C is [tex]3.35*10^5J/Kg[/tex]

Our values are given as,

m=53 kg

[tex]C=4180 J/kg\°C[/tex]

[tex]\Delta T=10-(-1)=11[/tex]

Replacing we have,

[tex]Q=53*4180*6+53*3.35*10^5[/tex]

[tex]Q = 19084240J[/tex]

Power can be defined as

[tex]P = \frac{Q}{t}[/tex]

Re-arrange to find t,

[tex]t = \frac{Q}{P}[/tex]

[tex]t = \frac{19084240}{1200}[/tex]

[tex]t = 15903.53s \approx 265 min \aprox 4.41h[/tex]

Therefore the time interval is 4.41h

The question revolves around the principle of thermal energy transfer and the specific heat of water. The water’s high specific heat absorbs a large amount of heat from the cellar and hence delays the freezing of the fruit by an additional 0.57 hours.

The subject of this question involves the principles of thermal Energy and heat transfer, specifically regarding the specific heat of water. The water in the tub has the ability to retain heat and slow down the cooling process of the space due to water's high specific heat capacity which is 4180 J/kg⋅°C. This means that water can absorb a lot of heat before its temperature rises.

Firstly, we need to calculate how much heat is the water in the tub able to absorb before its temperature reaches -1 °C (the freezing point of the fruit). This can be calculated with the formula for heat transfer Q = mcΔT, where m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. In this case, ΔT is 10 - (-1) = 11 °C, so Q = 53 kg * 4180 J/kg⋅°C * 11 °C = 2,444,840 J of heat.

The cellar loses heat at a rate of 1200 J/s, which means to lose 2,444,840 J of heat it would require 2,444,840 J / 1200 J/s = 2,037.3 seconds, which is approximately 0.57 hours. So, the presence of the water in the tub delays the freezing of the fruit by an additional 0.57 hours.

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To develop this problem it is necessary to use the continuity equations and Bernoullie's theorem.

It is known from Bernoullie's theorem that

[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

Where

P = Pressure

g = Gravity

h= Height

v = Velocity

[tex]\rho[/tex] = Density

On the other hand we have that the continuity equation is given by

[tex]A_1v_1 = A_2 v_2[/tex]

Where A is the Cross-sectional area and v the velocity.

For our values we know that

[tex]A_1v_1 = A_2 v_2[/tex]

[tex](\frac{\pi d_1^2}{4})v_1 =(\frac{\pi d_2^2}{4})v_2[/tex]

[tex]d_1^2v_1=d_2^2v_2[/tex]

[tex](11.5cm)^2(9.39)=(15.3)^2v_2[/tex]

[tex]v_2 = 5.305m/s[/tex]

Using Bernoulli's expression we can now find the pressure difference,

[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

[tex]P_1-P_2=-\rho gh_1-\frac{1}{2}\rho v_2^2 +\rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

[tex]P_1-P_2 = \rho g (h_1-h_2)+\frac{1}{2}\rho(v_1^2-v_2^2)[/tex]

[tex]P_1-P_2 = (1.3*10^3)(9.8)(9.59)+\frac{1}{2}(1.3*10^3)((9.39)^2-(5.305)^2)[/tex]

[tex]P_1-P_2 = 1.612*10^5Pa[/tex]

Answer: a)2.542cm

Explanation:

According to area expansivity which is defined as change in area per unit area for degree rise in kelvin.

Area expansivity= A2-A1/A1(¶2-¶1)

A2-A1 is change in area

¶2-¶1 is temperature change

A2 if final area

A1 is initial area

¶2 is final temp = 91°C

¶1 is initial temp= 26°C

coefficient of linear expansion for steel is 11 ✕ 10−6 (°C)−1.

Area of the spherical steel ball = Πd²/4

A1= Π×2.54²/4

A1 = 5.07cm²

Area expansivity = 2×linear expansion = 2×11 ✕ 10−6 (°C)−1.

= 22 ✕ 10−6 (°C)−1.

Substituting in the formula to get final area A2

22 ✕ 10−6 (°C)−1 = A2-5.07/5.07(91-26)

22 ✕ 10−6 (°C)−1 = A2-5.07/329.55

A2-5.07 = 0.0073

A2 = 0.0073+5.07

A2= 5.0073cm²

To get final diameter

A2=Πd²/4

5.0073=Πd²/4

20.309 = Πd²

d² = 20.309/Π

d²=6.46

d= √6.46

d= 2.542cm

Answer:

c. Both hit the ground with the same speed.

Explanation:

Given that speed of the both balls are same.

We know that kinetic energy given as

[tex]KE=\dfrac{1}{2}mv^2[/tex]

v=Speed  ,  m= mass

If mass of the the balls are same then their kinetic energy will be same.

The potential energy PE

PE= m g h

h =height

Given that h is same ,therefore their potential energy will be same.

But the time taken by both the balls will be different.

Therefore answer is c.

Both balls hit the ground with the same speed because they start with the same initial speed and experience the same influence of gravity. Hence, option (c) is the correct answer.

This question pertains to the principles of projectile motion. When two balls are thrown from the same height with the same initial speed, one straight up and one at a 45° angle, we can analyze their motion by breaking it into vertical and horizontal components.

Thus, the correct statement is (c) Both hit the ground with the same speed.

Final answer:

To calculate the sound intensity level, we use the formula I = (A² * ρ * v * f)/2. Plugging in the given values, the sound intensity level will be approximately 4.32 × 10⁻⁶ W/m².

Explanation:

To calculate the sound intensity level, we can use the formula:

I = (A² * ρ * v * f)/2

Where:

I is the sound intensity level,

A is the displacement amplitude,

ρ is the density of air,

v is the speed of sound,

f is the frequency of the sound wave.

Plugging in the given values, we have:

I = (2.00×10⁻⁸ m)² * 1.34 kg/m³ * 328 m/s * 530 Hz)/2

Simplifying the expression, the sound intensity level will be approximately 4.32 × 10⁻⁶ W/m².

Answer: 8.6 µm

Explanation:

At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:

Ey =Emax cos (kx-ωt)

So, we can write the following equality:

ω= 2.2 1014 rad/sec

The angular frequency and the linear frequency are related as follows:

f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec

In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.

The wavelength, speed and frequency, are related by this equation:

λ = c/f

λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.

The wavelength of the electromagnetic wave is 8.57 micrometers.

The wavelength of an electromagnetic wave can be calculated using the following equation:

λ = c / f

where:

λ is the wavelength

c is the speed of light in vacuum (3.0 × 10^8 m/s)

f is the frequency

The frequency of the electromagnetic wave can be calculated from the angular frequency ω using the following equation:

f = ω / 2π

where:

ω is the angular frequency (2.20 × 10^14 rad/s)

Plugging in the values for ω and c, we get:

f = (2.20 × 10^14 rad/s) / (2π)

f = 3.50 × 10^13 Hz

Now we can calculate the wavelength:

λ = (3.0 × 10^8 m/s) / (3.50 × 10^13 Hz)

λ = 8.57 × 10^-6 m

Therefore, the wavelength of the electromagnetic wave is 8.57 micrometers.

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To find the distance of closest approach between the alpha particle and the gold nucleus, set the potential energy equal to zero and solve for r.

In Rutherford's scattering experiments, alpha particles with a charge of +2e were fired at a gold foil. To find the distance of closest approach between the alpha particle and the gold nucleus, we need to calculate the potential energy at this point. Given that the initial kinetic energy of the alpha particle is 2.9 MeV, we can equate it to the electrical potential energy using the formula:

KE = PE

Since the alpha particle comes to rest when all its initial kinetic energy has been converted to electrical potential energy, we can set the potential energy equal to zero:

0 = k * (2e * 79e) / r

where k is the Coulomb constant and r is the distance of closest approach.

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Final answer:

To determine the distance of closest approach between an alpha particle and a gold nucleus with an initial kinetic energy of 2.9 MeV, we apply the conservation of energy principle and calculate the point where the kinetic energy is entirely converted into electrical potential energy.

Explanation:

The student is asking about the distance of closest approach between an alpha particle and the nucleus of a gold atom during Rutherford's scattering experiments when the initial kinetic energy of the alpha particle is 2.9 MeV. To find this distance, we use the concept of conservation of energy, where the initial kinetic energy of the alpha particle is completely converted into electrical potential energy at the point of closest approach. The formula for the electrical potential energy (U) at distance (r) is given by U = k*q1*q2/r, where k is Coulomb's constant, q1 and q2 are the charges of the alpha particle and gold nucleus, respectively, and r is the distance of closest approach.

The charge of an alpha particle is +2e (+2e), and the charge of a gold nucleus is +79e. Given that the initial kinetic energy (K) is 2.9 MeV, we can set K = U to solve for r. The calculation involves converting the given energy into Joules (1eV = 1.6x10^-19 J), and then using the values for k, q1 (+2e), and q2 (+79e) to determine r. Through this, we can derive the equation to calculate the distance of closest approach in femtometers (fm), noting that 1 fm = 1x10^-15 m.

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement  a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t  Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s  , t= 30s

θ = (58 rad/s)*(30)

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt  = 290 rad+ 1740 rad + 1201.42 rad

θt  = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

Answer:

a. Box 1

Explanation:

Hi there!

The momentum of the system box-ball is conserved in both cases because there is no external force applied on the system.

The momentum of the system is calculated as the sum of the momenta of each object that composes the system. The momentum is calculated as follows:

p = m · v

Where:

p =  momentum.

m = mass.

v = velocity.

Then, the momentum of the system before and after the collision will be:

System ball - box 1

initial momentum = final momentum

mb · vb + m1 · v1 = mb · vb´ + m1 · v1´

Where:

mb = mass of the ball.

vb = veloctiy of the ball.

m1 = mass of box 1.

v1 = velocity of box 1.

vb´ = final velocity of the ball.

v1´ = final velocity of box 1.  

Since the initial velocity of the box is zero:

mb · vb = mb · vb´ + m1 · v1´

Solving for v1´

mb · vb - mb · vb´ = m1 · v1´

mb · (vb - vb´) = m1 · v1´

mb · (vb - vb´) / m1 = v1´

Since vb´ is negative because the ball bounces back, then:

mb · (vb + vb´) / m1 = v1´

Now let´s express the momentum of the system ball - box 2

System ball -box 2

mb · vb + m2 · v2 = (mb + m2) · v2´

Since v2 = 0

mb · vb =  (mb + m2) · v2´

Solving for v2´:

mb · vb / (mb + m2) = v2´

Comparing the two expressions:

v2´ = mb · vb / (mb + m)

v1´ = mb · (vb + vb´) / m

In v1´ the numerator is greater than the numerator in v2´ because

vb + vb´> vb  

In v2´ the denominator is greater than the denominator in v1´ because

mb + m > m

then v1´ > v2´

Box 1 ends up moving faster than box 2

Answer:

a ) = 381.48 J

b )=  84.25 cm

Explanation:

Kinetic energy of the runner

= 1/2 m v²

= .5 x 66 x 3.4²

= 381.48 J

The final kinetic energy of the runner is zero .

Loss of mechanical energy

= 381.48 J

This  loss in  mechanical energy is due to action of frictional force .

b )

Let s be the distance of slide

deceleration due to frictional force

= μmg/m

.7 x 66 x 9.8 / 66

a = - 6.86 m s⁻¹

v² = u² - 2 a s

0 = 3.4² - 2x6.86 s

s = 3.4² / 2x6.86

= .8425 m

84.25 cm

Answer:

The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.

Explanation:

Given that,

Mass of block A = 6.00 kg

Mass of block B = 3.00 kg

Moment of inertia = 0.220 kg.m²

Radius = 0.120 m

Suppose we need to find the the magnitude of the linear acceleration of block A

Let a is the acceleration of the blocks.

Let [tex]T_{a}[/tex] and [tex]T_{b}[/tex] are the tension in the A and B cord.

According to figure,

We need to calculate the magnitude of the linear acceleration of block A

Net force acting on block A,

[tex]F_{A}=m_{A}g-T_{A}[/tex]

[tex]m_{A} a=m_{A}g-T_{A}[/tex]

[tex]T_{A}=m_{A}g-m_{A}a[/tex]...(I)

Net force acting on block B,

[tex]F_{B}=T_{B}-m_{B}g[/tex]

[tex]m_{B}a=T_{B}-m_{B}g[/tex]

[tex]T_{B}=m_{B}a+m_{B}g[/tex]...(II)

Net torque acting on pulley

[tex]T_{net}=I\times\alpha[/tex]

[tex]T_{A}r-T_{B}r=I\times \dfrac{a}{r}[/tex]

[tex]T_{A}-T_{B}=I\times\dfrac{a}{r^2}[/tex]

[tex]m_{A}g-m_{A}a-(m_{B}g+m_{B}a)=I\times\dfrac{a}{r^2}[/tex]

[tex]g(m_{A}-m_{B})-a(m_{A}+m_{B})=I\times\dfrac{a}{r^2}[/tex]

[tex]g(m_{A}-m_{B})=I\times\dfrac{a}{r^2}+a(m_{A}+m_{B})[/tex]

[tex]g(m_{A}-m_{B})=a(\dfrac{I}{r^2}+(m_{A}+m_{B}))[/tex]

[tex]a=\dfrac{g(m_{A}-m_{B})}{(\dfrac{I}{r^2}+(m_{A}+m_{B}))}[/tex]

Put the value into the formula

[tex]a=\dfrac{9.8\times(6.00-3.00)}{\dfrac{0.220}{(0.120)^2}+(6.00+3.00)}[/tex]

[tex]a=1.21\ m/s^2[/tex]

Hence, The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

[tex]M = \frac{gr^2}{G}[/tex]

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

[tex]g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m[/tex]

Replacing at this equation we have that

[tex]M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg[/tex]

The Volume of a Sphere is equal to

[tex]V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3[/tex]

Therefore using the relation between mass, volume and density we have that

[tex]\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3[/tex]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]

Here

P = Pressure

T = Temperature

[tex]\gamma =[/tex]The ratio of specific heats. For air normally is 1.4.

Our values are given as,

[tex]P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K[/tex]

Therefore replacing we have,

[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]

[tex](\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]

Solving for [tex]P_2,[/tex]

[tex]P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]

[tex]P_2 = 44.15Lbf/in^2[/tex]

Therefore the maximum theoretical pressure at the exit is [tex]44.15Lbf/in^2[/tex]

To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. Given the initial pressure and temperature, as well as the final temperature, we can calculate the maximum theoretical pressure using the equation p2,max = (p1 * T2) / T1.

To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. The ideal gas law is given by the equation pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for the pressure:

p = (nRT) / V

Since the volume and the number of moles are constant, we can rewrite the equation as:

p1 / p2 = T1 / T2

where p1 and T1 are the initial pressure and temperature, and p2 and T2 are the final pressure and temperature.

Substituting the given values, we have:

p2,max = (p1 * T2) / T1

Plugging in the values from the problem statement, we have:

p2,max = (15 lbf/in.2 * 275°F) / 80°F

p2,max = 51.56 lbf/in.2

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To solve for the normal force, the forces in the vertical direction are balanced, considering gravity and the y-component of force [tex]F_2[/tex] which is approximately 24.211 N. For acceleration, the net force in the horizontal direction is calculated by adding force [tex]F_1[/tex] and the x-component of [tex]F_2[/tex], and subtracting kinetic friction, then divided by the mass which gives [tex]23.14 m/s^2[/tex] approximately.

The problem given requires us to calculate the magnitude of the normal force and the acceleration of the block using Newton's second law and the concepts of kinetic friction.

Part (a)

Firstly, to find the normal force, [tex]F_N[/tex], we need to consider the forces in the vertical (y) direction. The gravitational force (weight) acts downward with a magnitude of m × g, and the vertical component of [tex]F_2[/tex] acts upward. Additionally, the normal force acts upward, balancing these forces.

Thus, we have:

[tex]F_N + F_{2y} - m \times g = 0[/tex]

Where:

[tex]F_N[/tex] is the normal force

[tex]F_{2y} = F_2[/tex] × sin(θ) (the vertical component of [tex]F_2[/tex])

m × g is the weight of the block, m is the mass and g is the acceleration due to gravity, which is approximately [tex]9.81 m/s^2[/tex]

Plugging in the known values:

[tex]F_N[/tex] + 12.4 × sin(30°) - 3.1 × 9.81 = 0

[tex]F_N[/tex] = 3.1 × 9.81 - 12.4 × 0.5

[tex]F_N[/tex] = 30.411 - 6.2

[tex]F_N[/tex] = 24.211 N (approximately)

Part (b)

For the acceleration, a, we consider the forces in the horizontal (x) direction. The net force is the sum of the horizontal component of [tex]F_2[/tex] and [tex]F_1[/tex] minus the kinetic friction force, [tex]F_k[/tex].

The kinetic friction force, [tex]F_k[/tex], is given by:

[tex]F_k = \mu_k \times F_N[/tex]

And hence:

[tex]F_{net}[/tex] = [tex]F_1 + F_{2x} - F_k[/tex]

Using [tex]F_N[/tex] from the first section to find [tex]F_k[/tex] and then [tex]F_{net}[/tex], we get:

[tex]F_{net}[/tex] = 65 + 12.4 × cos(30°) - 0.2 × 24.211

Using Newton's second law, a = [tex]F_{net}[/tex] / m:

a = (65 + 12.4 × 0.866 - 0.2 × 24.211) / 3.1

a = (65 + 10.7424 - 4.8422) / 3.1

a = 71.7424 / 3.1

a = [tex]23.14 m/s^2[/tex] (approximately)

Answer:

The mass of the plank is 33 kg.

Explanation:

Given that,

Length of plank = 4 m

Distance in left from the center= 0.5 m

Mass of child = 42 kg

Mass of other child = 31 kg

Distance in right from the center= 1 m

We can sum torques around the saw horse; each force generates a torque equal to force x distance from saw horse

We need to calculate the torque for first child

Using formula of torque

[tex]\tau =mg\times d[/tex]

Put the value into the formula

[tex]\tau=42\times9.8\times1.5[/tex]

[tex]\tau=617.4\ N-m[/tex]

We need to calculate the torque for second child

The child is 1.5 m from the saw horse

Using formula of torque

[tex]\tau' =mg\times d[/tex]

Put the value into the formula

[tex]\tau'=31\times9.8\times1.5[/tex]

[tex]\tau'=455.7\ N-m[/tex]

The weight of the plank produces a clockwise torque

[tex]\tau''=mg\times d[/tex]

[tex]\tau''=mg\times 0.5[/tex]

We need to calculate the mass of the plank

Using balance equation of torque

[tex]\tau=\tau''+\tau'[/tex]

Put the value into the formula

[tex]617.4=mg\times0.5+455.7[/tex]

[tex]m=\dfrac{617.4-455.7}{0.5\times9.8}[/tex]

[tex]m=33\ kg[/tex]

Hence, The mass of the plank is 33 kg.

The mass of the plank is found to be 29 kg.

To solve for the mass of the plank, we need to use the principles of static equilibrium. The seesaw must be balanced, meaning the sum of moments around the pivot (saw horse) must be zero.

Given -

Length of the plank = 4 m

Pivot position (0.5 m to the left of the center) = 2-0.5 m from the left end = 1.5 m

Mass of the first child, m1 = 42 kg (at the left end, 1.5 m from pivot)

Mass of the second child, m2 = 31 kg (1 m to the right of the center, 1.5+1 m = 2.5 m from the pivot)

Equation of moments about the pivot (taking counterclockwise as positive):

Moment of m1 = 42 kg * 1.5 m (clockwise) = -42 * 1.5 Nm

Moment of m2 = 31 kg * 2.5 m (counterclockwise) = 31 * 2.5 Nm

Moment of the plank's weight (assuming it acts at its center, which is 0.5 m from the pivot):

Moment of plank = M * 0.5 m (clockwise) = -M * 0.5 Nm

Setting the sum of moments to zero for equilibrium:

31 * 2.5 - 42 * 1.5 - M * 0.5 = 0

Solving for M:

31 * 2.5 - 42 * 1.5 = M * 0.5

77.5 - 63 = 0.5M

14.5 = 0.5M

M = 14.5 / 0.5 = 29 kg

Therefore, the mass of the plank is 29 kg.