A box with a frontal area of 2.5 ft2 and with a drag coefficient of CD =0.9 is fastened on the roof of your car. What is the aerodynamic drag force in pounds due to the box alone at a velocity of 88 ft/sec?

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

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:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

[tex]\mu[/tex] = Coefficient of friction

[tex]\theta[/tex] = Slope

From Newton's second law

[tex]mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655[/tex]

Applying [tex]\mu[/tex] to the above equation and [tex]\theta=10^{\circ}[/tex]

[tex]mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2[/tex]

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

[tex]\Delta E = E_0 -W_f[/tex]

Where

[tex]E_0 =[/tex]Initial Energy

[tex]W_f =[/tex] Energy loses

Replacing we have that

[tex]\Delta E = 1.6-0.8[/tex]

[tex]\Delta E = 0.8eV[/tex]

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

Answer:

E)  I = 18.4 N.s

Explanation:

For this exercise let's use momentum momentum

     I = Δp = [tex]p_{f}[/tex]- p₀

The energy of the stone is only kinetic

    K = ½ m v²

The initial energy is Ko and the final is 70% Ko

     [tex]K_{f}[/tex] = 0.70 K₀

energy equation

     [tex]K_{f}[/tex] = 0.7 ½ m v₀²

You can also write

     [tex]K_{f}[/tex] = ½ m vf²

   ½ m vf² = ½ m (0.7 v₀²)

   [tex]v_{f}[/tex] = v₀ √ 0.7

Now we can calculate and imposed

     I = m (-vo √0.7) - m vo

     I = m vo (1 +√0.7

     I = 0.5000 20.0 (1.8366)

     I = 18.4 N.s

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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Answer:

4.22 mm

Explanation:

E = Young’s modulus for steel = 210 GPa (generally)

[tex]\Delta L[/tex] = Change in length = 3 mm

[tex]L_0[/tex] = Original length = 4 m

A = Area of cable

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of cable

d = Diameter = 2r

m = Mass of chandelier = 226 kg

[tex]\epsilon[/tex] = Longitudinal strain = [tex]\frac{\Delta L}{L_0}[/tex]

Uniaxial stress is given by

[tex]\sigma=E\epsilon\\\Rightarrow \sigma=210\times 10^9 \frac{3\times 10^{-3}}{4}\\\Rightarrow \sigma=157500000\ Pa[/tex]

[tex]\sigma=\frac{F}{A}\\\Rightarrow \sigma=\frac{mg}{\pi r^2}\\\Rightarrow 157500000=\frac{226\times 9.81}{\pi r^2}\\\Rightarrow r=\sqrt{\frac{226\times 9.81}{157500000\times \pi}}\\\Rightarrow r=0.00211\ m\\\Rightarrow d=2r\\\Rightarrow d=2\times 2.11=4.22\ mm[/tex]

The diameter of the cable is 4.22 mm

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 with Explanation:

We are given that

Electric field vector=[tex]\vec{E}=E\hat{i}[/tex]

Magnetic field vector=[tex]\vec{B}=-B\hat{j}[/tex]

We have to find the direction of propagation of the wave.

We know that  

The direction of propagation of wave=[tex]\vec{E}\times \vec{B}[/tex]

The direction of propagation of wave=[tex]\hat{i}\times (-\hat{j})=-\hat{k}[/tex]

Because ([tex]\hat{i}\times \hat{j}=\hat{k}[/tex])

The  wave propagate in -z direction.

Therefore, the component of the vector is (0,0,-1).

Hence, the direction of the propagation of wave is - z direction and component of direction of propagation vector is (0,0,-1).

The direction of propagation of an Electromagnetic wave is perpendicular to the electric and magnetic fields of the wave. In this case, the fields are along the x and negative y axes, so by applying the cross product, the propagation is in the positive z-direction, hence the unit vector would be 0,0,1.

In physics, an Electromagnetic wave has an electric field (E) and magnetic field (B) that are perpendicular to each other and the direction of propagation. These fields are also in phase with each other. Due to these properties, the direction of propagation can be calculated using the cross-product of vectors E and B.

By following the right-hand rule, if you imagine your fingers to curl from E towards B, your thumb will point in the direction of wave propagation. In this case, E = Ei (in the direction of the x-axis) and B = -Bj (in the direction of the negative y-axis).

Therefore, the cross-product E x B = E(-B)k^ (in the direction of the z-axis). Hence, the direction of the propagation Vector, P⃗, will be in the direction of the positive z-axis. For a unit vector, this would simply be 0,0,1.

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Answer:

66.98 db

Explanation:

We know that

[tex]L_T=L_S+10log(n)[/tex]

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

[tex]L_T=60+10 log(5)[/tex]

= 66.98 db

The combined sound level of five people talking at 60 dB each, given the logarithmic nature of the decibel scale, would be approximately 73 dB.

The sound level in decibels (dB), which is a measure of the sound's loudness, is calculated using the formula: B (dB) = 10 log( I / Io ) where B is the decibel level, I is the sound intensity, and Io is the reference intensity (10^-12 W/m² in this case). If one person talking is 60 dB then five people talking would add up logarithmically, not linearly, because the decibel scale is a logarithmic scale.

The formula to calculate the combined sound level of multiple sources is: Total dB = 10 log10 (10^(dB1/10) + 10^(dB2/10) + …). In our case, it would be Total dB = 10 log10 (5*10^(60/10)), which is approximately 73 dB.

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To solve this problem it is necessary to apply the concepts related to heat exchange and Entropy.

The temperature and mass remain constant, therefore the entropy values will be the only ones to change.

Of the three elements given their entropy values are given by

[tex]\Delta S_{iron}=0.45J/gmC[/tex]

[tex]\Delta S_{glass}=0.8J/gmC[/tex]

[tex]\Delta S_{water}= 4.186J/gmC[/tex]

Part A) From thermodynamic theory we know that temperature is inversely proportional to entropy

[tex]T = \frac{E}{\Delta S} \rightarrow[/tex] Energy remains constant

[tex]\Delta t \propto \frac{1}{S}[/tex]

Therefore the order would be

Lowest Temperature= Water

Medium Temperature= Glass

Highest Temperature=Iron

Part B) In the case of Energy the opposite happens because it is proportional to the entropy, then

[tex]E = T(\Delta S) \rightarrow[/tex]Temperature is constant

[tex]E \propto S[/tex]

Lowest Energy = Iron

Medium Energy = Glass

Highest Energy = Water

The temperature increase brought by a given amount of energy input depends on the material's specific heat capacity. Given a certain amount of energy, iron will heat up the most due to its low specific heat capacity. But if a certain temperature increase is desired, water, having high specific heat, will require the most energy.

The ranking of the samples after adding 100 J of energy can be determined by comparing their specific heat capacities. The specific heat capacity is the amount of heat necessary to raise the temperature of a substance by one degree Celsius.

Iron has a lower specific heat capacity than glass and water, meaning it will heat up more quickly when the same amount of energy is added. Thus, providing the same amount of energy (100 J) to each sample would result in the temperature of the iron increasing to the greatest extent, followed by the glass, and then the water.

Conversely, when each sample is to increase in temperature by the same amount (20°C), the amount of energy required will also be determined by the specific heat capacities. This time, however, the substance with the higher specific heat capacity will require more energy to achieve this temperature increase. Therefore, the water will require the most energy, followed by the glass, and then the iron.

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Answer:

[tex]f=145.29Hz[/tex]

Explanation:

The centripetal force is given by:

[tex]F_c=ma_c(1)[/tex]

Here m is the body's mass in which the force is acting and [tex]a_c[/tex] is the centripetal acceleration:

[tex]a_c=\frac{v^2}{r}(2)[/tex]

Here v is the speed of the body and r its radius. The speed is given by:

[tex]v=2\pi fr(3)[/tex]

Replacing (3) in (2):

[tex]a_c=4\pi^2f^2r(4)[/tex]

Replacing (4) in (1) and solving for f:

[tex]F_c=m4\pi^2 f^2r\\\\f=\sqrt{\frac{F_c}{4m\pi^2r}}\\f=\sqrt{\frac{4*10^{-11}N}{4(3*10^{-16}kg)\pi^2(16*10^{-2}m)}}\\f=145.29Hz[/tex]

Final answer:

To find the revolutions per second for a centrifuge, the known force acting on a red blood cell is used with the mass and radius of the centrifuge to calculate the angular velocity, which is then converted to revolutions per second.

Explanation:

To determine the revolutions per second at which a centrifuge should be operated to exert a specified force on a red blood cell, we can make use of the relationship between centrifugal force and centripetal acceleration. The force experienced by the red blood cell can be expressed by Newton's second law for circular motion: F = m\(a_{c}\), where F is the force, m is the mass of the red blood cell, and \(a_{c}\) is the centripetal acceleration.

The centripetal acceleration \(a_{c}\) is also related to the angular velocity \(\omega\) of the rotating body by the formula \(a_{c} = \omega^{2}r\), where r is the radius of the centrifuge. Solving for the angular velocity gives us \(\omega = \sqrt{\frac{F}{mr}}\). Once we have \(\omega\), we can convert it to revolutions per second by dividing by 2\(\pi\) (since one revolution is 2\(\pi\) radians).

For the given numbers, F = 4.0 \(\times\) 10^{-11} N, m = 3.0 \(\times\) 10^{-16} kg, and r = 16.0 cm = 0.16 m. We can calculate \(\omega\), and subsequently the number of revolutions per second the centrifuge should be operated at to achieve the necessary centrifugal force for sedimentation of the blood cells. The extraordinarily high centripetal acceleration makes such devices known as ultracentrifuges.

The first disk's kinetic energy before it is dropped is 18.75 J. The final angular velocity of the combined disks after they stick together is 3.33 rad/s. The final kinetic energy of the combined disks is 24.97 J.

Given a solid disk of mass 3 kg, radius 0.5 m, and initial angular velocity 10 rad/sec, and a second disk with the same radius but with double the mass (6 kg), which is initially at rest, we need to find the following:

(a) The kinetic energy (KE) of the first disk before it is dropped is calculated using the formula for rotational kinetic energy, KE = (1/2)Iω², where I is the moment of inertia for a solid disk (I = (1/2)mr²) and ω is the angular velocity. Therefore, KE = (1/2)((1/2)mr²)ω² = (1/2)((1/2)(3 kg)(0.5 m)²)(10 rad/s)² = 18.75 J.

(b) To find the final angular velocity (ω'), we conserve angular momentum since no external torques are acting. L_initial = L_final, which means Iω = (I1 + I2)ω'. Calculating, we get (1/2)(3 kg)(0.5 m)²(10 rad/s) = ((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)ω', solving for ω' gives us 3.33 rad/s.

(c) The final kinetic energy (KE_final) is calculated with the combined moment of inertia and the final angular velocity using KE_final = (1/2)(I1 + I2)ω'^2. Using the values from (b), we have KE_final = (1/2)((1/2)(3 kg)(0.5 m)² + (1/2)(6 kg)(0.5 m)²)(3.33 rad/s)² = 24.97 J.

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

[tex]|\Delta E_{ij}| = |E_i-E_j|[/tex]

Our states are given by

[tex]E_1 = -11.7eV[/tex]

[tex]E_2 = -4.2eV[/tex]

[tex]E_3 = -3.3eV[/tex]

In this way the energy balance for the states would be given by,

[tex]|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\[/tex]

[tex]|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV[/tex]

[tex]|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV[/tex]

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

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: 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:

5.55935324 m

Explanation:

r = Radius = 4 m

h = Height = 6 m

Frequency is

[tex]f=\frac{1}{T}\\\Rightarrow f=\frac{1}{5}\ Hz[/tex]

Angular speed is given by

[tex]\omega=2\pi f\\\Rightarrow \omega=\frac{2}{5}\pi[/tex]

Tangential velocity of the hat is given by

[tex]v=r\omega\\\Rightarrow v=4\times\frac{2}{5}\pi\\\Rightarrow v=5.02654\ m/s[/tex]

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

From equation of motion

[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 6=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{6\times 2}{9.81}}\\\Rightarrow t=1.106\ s[/tex]

Distance = Speed×Time

[tex]Distance=5.02654\times 1.106=5.55935324\ m[/tex]

The hat will land 5.55935324 m away from the point of release

David's hat would travel in a straight line relative to the Earth and land to the south of the point where it fell off due to the carousel's counterclockwise rotation and the principles of projectile motion.

When we consider David's hat falling off while he is riding on a flying carousel, we have to address the physical principles that govern the motion of the hat after it leaves his head. As David's hat falls off on the west side of the carousel, it would have initially the same horizontal velocity as the carousel at that point. However, since the carousel is rotating counterclockwise, and assuming there is no significant air resistance, the hat would move in a straight line relative to the Earth. This means it would land to the south of the point where it fell off, as the carousel would continue to rotate after the hat has been released. This situation applies principles of projectile motion and inertial frames.

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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:

v = 31.3 m / s

Explanation:

The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.

Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane

Highest point

       Em₀ = U = m g y

Lowest point

     [tex]Em_{f}[/tex] = K = ½ m v²

As there is no friction, mechanical energy is conserved

       Em₀ = [tex]Em_{f}[/tex]

       m g y = ½ m v²

       v = √ 2 g y

Where we can use trigonometry to find and

       sin 30 = y / L

       y = L sin 30

Let's replace

      v = RA (2 g L sin 30)

Let's calculate

      v = RA (2 9.8 100.0 sin30)

      v = 31.3 m / s

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.

Vertical distance, when the ball drops as it moves from the pitcher to the catcher is 1.048 meters.

The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.

The second equation of the motion for distance can be given as,

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Here, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the acceleration of the body and [tex]t[/tex] is the time taken by it.

Total time taken by the ball to reach to the catcher is the ratio of horizontal distance traveled by it to the speed of the ball.

As the speed of the ball is 40.2 m/s and the distance of the pitcher and the catcher is 18.6 meters. Thus the time of flight is,

[tex]t=\dfrac{18.6}{40.2}\\t=0.462\rm s[/tex]

Thus, total time taken by the ball to reach to the catcher is 0.462 seconds.

As the initial velocity of the ball is zero. Thus by the second equation of motion,

[tex]h=0+\dfrac{1}{2}(9.81)(0.462)^2\\h=1.048\rm m[/tex]

Thus, the vertical distance covered when the ball drops as it moves from the pitcher to the catcher is 1.048 meters.

Learn more about the equation of motion here;

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The baseball dropped approximately 1.04 meters vertically while it traveled horizontally to the catcher.

The subject of this question is Physics, specifically related to the concept of projectile motion. To find the vertical distance that the ball drops, we'll use the principles of kinematics. In this case, the problem describes a horizontal launch, and we can find the vertical distance using the equation for distance in free fall under gravity: d = 0.5*gt^2, where g = 9.8 m/s^2 is the acceleration due to gravity. The time t can be found from the horizontal motion, by dividing the horizontal distance by the horizontal speed: t = 18.6 m / 40.2 m/s = 0.46 s. Using this time, we compute the vertical drop: d = 0.5*9.8 m/s^2*(0.46 s)^2 = 1.04 m. So, the ball drops by approximately 1.04 m while it travels horizontally to the catcher.

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Answer:

a. The human body has nearly the same density as salt water after exhaling.

b. The human body will always float in the Dead Sea.

Explanation:

According to the concept of floating on the basis of  density, any body that is put in a fluid of density greater than its own density will always float due to the force of buoyancy from the liquid.

[tex]\rho_f.V_s=\rho_o.V_o[/tex]

where:

[tex]\rho_f=[/tex] density of the fluid

[tex]\rho_o=[/tex] density of the object

[tex]V_s=[/tex] volume of the object submerged in the fluid

[tex]V_o=[/tex] total volume of the object

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]

Answer:

Crushing force will be [tex]2.026\times 10^5N[/tex]

Explanation:

We have given that atmospheric pressure [tex]P=1.013\times 10^5N/m^2[/tex]

Area is given as [tex]A=2m^2[/tex]

We have to find the force

We know that force is given by

[tex]Force=pressure\times area=1.013\times 10^5\times 2=2.026\times 10^5N[/tex]

The reason that we do not get crushed is because the force is distributed  over our bodies, and because we have air inside our bodies pushing outwards (the air in your blood,lungs, etc.). The pressure inside of our bodies keeps the air outside from crushing it.

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) 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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Answer:

122.6 nPa

Explanation:

Here ,

Intensity of incident light , I = 36.8 W/m^2

angle with area, θ = 0 degree

for a perfect absorber ,

radiation pressure = [tex]\frac{Icos\theta}{c}[/tex]

I=  Intensity of incident light, c= speed of light

radiation pressure =  [tex]\frac{36.8\times cos0}{3\times10^8}[/tex]

radiation pressure = 12.26×10^-8 Pa

radiation pressure = 122.6 nPa

the radiation pressure acting on the 122.6 nPa

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:

The distance between the bright fringes is 2.19 mm

Explanation:

Distance of the screen from the slits, d = 1.9 m

Slit width, w = 0.11 mm = [tex]0.11\times 10^{- 3}\ m[/tex]

Wavelength, [tex]\lambda = 695\ nm[/tex]

Wavelength, [tex]\lambda' = 407\ nm[/tex]

To calculate the distance between the second order bright fringe:

[tex]y_{n} = \frac{n\lambda d}{w}[/tex]

[tex]y'_{3} = \frac{n\lambda' d}{w} = \frac{3\times 695\times 10^{- 9}\times 1.9}{0.11\times 10^{- 3}} = 0.036\ m[/tex]

[tex]y_{2} = \frac{2\times 407\times 10^{- 9}\times 1.9}{0.11\times 10^{- 3}} = 0.014\ m[/tex]

Distance, |x| = [tex]y'_{3} - y_{2}[/tex]

|x| = [tex]0.036 - 0.0141 = 0.0219 m = 2.19\ mm[/tex]