A high-wire walker always attempts to keep his center of mass (com) over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of 71.0 kg and a rotational inertia of 14.0 kg · m² about the wire. What is the magnitude of his angular acceleration about the wire if his com is 5.0 cm to the right of the wire under the following conditions?
(a) he carries no pole and
(b) the 14.0 kg pole he carries has its com 10 cm to the left of the wire?

Answer:

Q=1.42 C

Explanation:

Given that

[tex]I=I_oe^{\dfrac{-t}{\tau}}[/tex]

When t= 0 ,Io = 3 A

τ = Time constant = 0.5 s

[tex]I=3e^{-\dfrac{t}{\tau}}[/tex]

We know that

[tex]I=\dfrac{dQ}{dt}[/tex]

Q=Charge ,I =Current

Q = ∫I.dt

Given that

t= 0 to t= 3τ= 1.5 s

The charge Q

[tex]Q=\int_{0}^{1.5}3e^{-\dfrac{t}{0.5}}dt[/tex]

Q=1.42 C

Therefore charge flow conductor will be 1.42 C.

When the mass of Object 1 is reduced to one-fourth, the mass of Object 2 is tripled, and the separating distance is halved, the new gravitational force between these objects will be three times the original force. Thus, the new gravitational force will be 216.0 units.

The question asks us to determine the new gravitational force between two objects given that the mass of Object 1 is reduced to one-fourth of its original mass (so M1 becomes 0.25M1), the mass of Object 2 is tripled (so M2 becomes 3M2), and the distance separating the objects is halved (so R becomes 0.5R). Since gravitational force is calculated using the equation F = G*(M1*M2)/(R^2), where G is the universal gravitational constant, we can plug our new values into this equation.

So, the new gravitational force F' = G*(0.25M1 * 3M2)/(0.5R)^2. This simplifies further to F' = G*(0.75M1M2)/(0.25R^2) = 3 *(G*M1M2/R^2) = 3 * original force. Therefore, if the original force was 72.0 units, the new gravitational force is 3 * 72.0 = 216.0 units.

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To find the new gravitational force between the objects, we can use Newton's Law of Universal Gravitation.

The new gravitational force can be calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let's denote the original mass of Object 1 as m1, the original mass of Object 2 as m2, and the original distance separating them as d. The original gravitational force can be represented as:

F1 = G * (m1 * m2) / d²

After the given changes, the new mass of Object 1 is 1/4 * m1, the new mass of Object 2 is 3 * m2, and the new distance is 1/2 * d. Plugging these values into the equation, we find the new gravitational force:

F2 = G * ((1/4 * m1) * (3 * m2)) / (1/2 * d)²

Simplifying this equation will give us the answer.

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Answer:

[tex]c_m =272.365\frac{J}{KgK}[/tex]

Explanation:

1) Notation and data given

[tex]m_c[/tex]= mass of the container = 3.9 kg

[tex]m_w[/tex]= mass of the water inside of the container= 11 kg

[tex]m_m[/tex]=mass of the metal= 2 kg

[tex]T_{im}=189\degree C[/tex] initital temperature of the metal

[tex]T_{iw}=16\degree C[/tex] initital temperature of the water

[tex]T_{ic}=16\degree C[/tex] initital temperature of the container

[tex]T_{f}=18\degree C[/tex] final equilibrium temperature

2) Concepts and formulas

Since the inital temperature for the pice of metal is higher than the temperature for the container and the water inside, the piece of metal will transfer heat, and this transferred heat is the same amount absorbed by the container and the water reaching the equilibrium. So then we have this formula:

[tex]Q_m =Q_w + Q_c[/tex]

We don't have any change of phase so then we just have the presence of sensible heat, and using this we got that:

[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_c \Delta T[/tex]

On this case the container is made by the same metal so then [tex]c_c=c_m[/tex]

[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_m \Delta T[/tex]

And solving for [tex]c_m[/tex] from the last expression we got:

[tex]m_p c_m \Delta T_m -m_c c_m \Delta T_c =m_w c_w \Delta T[/tex]

[tex]c_m [m_p \Delta T_m -m_c \Delta T_c] =m_w c_w \Delta T[/tex]

[tex]c_m=\frac{m_w c_w \Delta T}{m_p \Delta T_m -m_c \Delta T_c}[/tex]

Replacing the values we have:

[tex]c_m=\frac{11kg(4187\frac{J}{KgK}(18\degree C-16\degree C)}{2kg(189\degree C -16 \degree C) -3.9Kg(18\degree C -16\degree C)}= 272.365\frac{J}{Kg K}[/tex]

Answer:

Part a)

[tex]E = 1.11 \times 10^5 N/C[/tex]

Part 2)

[tex]F = 1.78 \times 10^{-14} N[/tex]

Part 3)

[tex]W = 5 \times 10^{-17} J[/tex]

Explanation:

Part 1)

As we know that electric field and potential difference related to each other as

[tex]E = \frac{\Delta V}{x}[/tex]

so we will have

[tex]\Delta V = 614 V[/tex]

[tex]x = 5.51 mm[/tex]

so we have

[tex]E = \frac{614}{5.51 \times 10^{-3}}[/tex]

[tex]E = 1.11 \times 10^5 N/C[/tex]

Part 2)

Charge of an electron

[tex]e = 1.6 \times 10^{-19} C[/tex]

now force is given as

[tex]F = qE[/tex]

[tex]F = (1.6 \times 10^{-19})(1.11 \times 10^5)[/tex]

[tex]F = 1.78 \times 10^{-14} N[/tex]

Part 3)

Work done to move the electron

[tex]W = F.d[/tex]

[tex]W = (1.78 \times 10^{-14})(5.51 - 2.7) \times 10^{-3}[/tex]

[tex]W = 5 \times 10^{-17} J[/tex]

Answer:

Option (E)

Explanation:

The Rock glaciers are a type of landform that is formed due to the downward motion of angular blocks under the influence of gravity and the continuous cycle of freeze and thaw method. It mostly occurs in the steep slopes where the soil particles and the rock materials are prone to landslide. Here, the materials are mixed and covered with ice. This also occurs in the region where the permafrost layer undergoes creeping.

This downward motion of ice erodes the soil particles and carries with them further, from the site of talus cone or moraines and extends outward.

Thus, the correct answer is option (E).

Answer:

one of it is e

Explanation:

Answer:

360N

Explanation:

The frictional force = centripetal force = mass * centripetal acceleration = ma

Since centripetal acceleration = V^2/ r

Frictional force = mv^2 / r

Substitute the values of m = 80kg, radius = 2m and tangential speed = 3m/s

Frictional force = 80 * (3)^2/ 2 = 360N

The force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.

To calculate the force of friction that keeps an 80-kg person sitting on the edge of a horizontal rotating platform, we need to find the centripetal force required to keep the person moving in a circular path. The centripetal force is provided by the frictional force between the person and the platform.

 The formula for centripetal force [tex](\(F_c\))[/tex] is given by:

[tex]\[ F_c = \frac{mv^2}{r} \][/tex]

Let's plug in the values:

[tex]\[ F_c = \frac{(80\text{ kg})(3\text{ m/s})^2}{2\text{ m}} \][/tex]

[tex]\[ F_c = \frac{(80\text{ kg})(9\text{ m}^2/\text{s}^2)}{2\text{ m}} \][/tex]

[tex]\[ F_c = \frac{720\text{ kg}\cdot\text{m}/\text{s}^2}{2} \][/tex]

[tex]\[ F_c = 360\text{ N} \][/tex]

 Therefore, the force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.

 The answer is: [tex]360\text{ N}[/tex]

Final answer:

The gauge pressure reading stays the same when a steel cylinder is moved to the top of a mountain because it measures pressure relative to the external atmospheric pressure, which does not influence the pressure inside the cylinder.

Explanation:

The correct statement for the gauge that reads gauge pressure when a steel cylinder is brought to the top of a mountain is that the pressure reading stays the same. Gauge pressure is the pressure relative to atmospheric pressure, which means it measures the excess pressure in the tank over the external atmospheric pressure. Since the pressure inside the steel cylinder is not dependent on external atmospheric pressure changes, the reading on the gauge pressure meter would remain constant, even if the cylinder is taken to a different altitude.

Answer:

A. 50.7 J

B. 33.18 J

Explanation:

At the start of the motion the kinetic energy of the ball would be

[tex]E_k = \frac{mv^2}{2} = \frac{0.15*26^2}{2} = 50.7  J[/tex]

When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start

[tex]v_h = vcos(36^o) = 26*0.81 = 21.03 m/s[/tex]

So the kinetic energy at this point is

[tex]E = \frac{mv_h^2}{2} = \frac{0.15*21.03^2}{2} = 33.18J[/tex]

Given the data from the question, the kinetic energy at the start and at the highest point are:

A. The kinetic energy at the start of the motion is 50.7 J

B. The kinetic energy at the highest point is 0 J

This is the energy possessed by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

KE = ½mv²

KE = ½ × 0.15 × 26²

KE = 50.7 J

KE = ½mv²

KE = ½ × 0.15 × 0

KE = 0 J

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Answer:

Layer of the interior of the Earth that extends approximately between 50 and 150 km deep, formed mainly by partially molten plastic rocks that can deform: the asthenosphere is located between the lithosphere and the mesosphere.

Explanation:

The asthenosphere is composed of ductile silicate materials, in solid state and partially or totally semi-molten (according to their depth and / or proximity to magma bags), which allow continental drift and isostasis. On it the tectonic plates move.

Characteristics of the asthenosphere

The asthenosphere is a part of the upper mantle, just below the lithosphere that is involved in tectonic plate movements and isostatic adjustments.

Despite its heat, the pressure keeps it like plastic and has a relatively low density. Seismic waves pass relatively slowly through the asthenosphere, compared to the mantle that covers the lithosphere, which is why it has been called the low-speed zone.

This was the observation that originally alerted the seismologists to their presence and gave some information about their physical properties, such as the speed of the seismic waves that decreases as stiffness decreases. Under the thin oceanic plates the asthenosphere is generally much closer to the surface of the seabed, and rises a few kilometers from the ocean floor.

Due to the temperature and pressure conditions in the asthenosphere, the rock becomes ductile, moving at a deformation rate that is measured in cm / year at linear distances that finally reach thousands of kilometers. In this way, it flows like a convection current, radiating heat outward from inside the Earth. Above the asthenosphere, at the same rate of deformation, the rock behaves elastically and, if fragile, can break, causing failures. The rigid lithosphere is believed to "float" over the asthenosphere from which it flows slowly, creating the movement of tectonic plates.

Answer:

The conistency is solid layer of rock

Explanation:

Explanation:

It is given that,

Velocity of the electron, [tex]v=(2\times 10^6i+3\times 10^6j)\ m/s[/tex]

Magnetic field, [tex]B=(0.030i-0.15j)\ T[/tex]

Charge of electron, [tex]q_e=-1.6\times 10^{-19}\ C[/tex]

(a) Let [tex]F_e[/tex] is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

[tex]F_e=q_e(v\times B)[/tex]

[tex]F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)][/tex]

[tex]F_e=-1.6\times 10^{-19}\times (-390000)(k)[/tex]

[tex]F_e=6.24\times 10^{-14}k\ N[/tex]

(b) The charge of electron, [tex]q_p=1.6\times 10^{-19}\ C[/tex]

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

Answer:

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force  + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} ,  mass = density x volume

        where density is the density of air for the buoyant force

        buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

        force by helium =  0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

        force of its weight = 0.0125 x 9.8 = 0.1225 N

         the force  of weight of rubber and of helium act downwards, so they      

          carry a negative sign.

Calculating the tension on a string tied to a helium-filled balloon involves principles of buoyancy and weight. The difference between the buoyant force exerted by the displaced air and the sum of the weight of the empty balloon and weight of the helium gives the tension on the line.

The tension on the vertical line tied to the helium-filled balloon can be found using Archimedes' Principle and Buoyancy. First, the volume (V) of the helium-filled balloon can be calculated using the formula for the volume of a sphere; V = 4/3 x π x (0.498 m)³ = ~0.516 m³. The mass (m) of the helium in the balloon is calculated by multiplying this volume by the density of helium, which is m = 0.516 m³ x 0.181 kg/m³ = ~0.093 kg.

Next, the weight (w) of air displaced by the balloon is calculated. This weight is equal to the product of the volume of the balloon, the density of the air, and gravitational acceleration; w = 0.516 m³ x 1.29 kg/m³ x 9.8 m/s² = ~6.5 N. This is the buoyant force that lifts the balloon.

The weight (W) of the filled balloon is the sum of the weight of the empty balloon and the helium; W = (0.0125 kg + 0.093 kg) x 9.8 m/s² = ~1.03 N. The tension (T) in the line is then the difference between the buoyant force and this weight (i.e., T = 6.5 N - 1.03 N =  ~5.47 N).

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Answer:

Average force on the football = 168.82 N

Explanation:

Force = Mass x Acceleration

F = ma

Mass, m = 0.41 kg

We have equation of motion, v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 21 m/s

Time, t = 0.051 s

Substituting

                  21 = 0 + a x 0.051

                     a = 411.76 m/s²

Substituting in force equation,

                   F = ma = 0.41 x 411.76 = 168.82 N

Average force on the football = 168.82 N

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

[tex]W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m[/tex]

Answer:-81.29 J

Explanation:

Given

mass of ball [tex]m=2.35 kg[/tex]

Length of string [tex]L=3.53 m[/tex]

height of Room [tex]h=5.03 m[/tex]

Gravitational Potential Energy is given by

[tex]P.E.=mgh[/tex]

where h=distance between datum and object

here Reference is ceiling

therefore [tex]h=-3.53 m[/tex]

Potential Energy of ball w.r.t ceiling

[tex]P.E.=2.35\times 9.8\times (-3.53)=-81.29 J[/tex]

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling    

Answer:

Radiated power will be [tex]675241.7175watt[/tex]

Explanation:

We have given emissivity [tex]\epsilon =0.9[/tex]

Stove temperature T = 455 K

Room temperature [tex]T_C=295K[/tex]

We know the Stephan's constant [tex]\sigma =5.67\times 10^{-6}w/m^2K^4[/tex]

We know that radiated power is given by [tex]P=\epsilon \sigma A(T^4-T_C^4)=0.9\times5.67\times 10^{-6}\times  3.75\times (455^4-295^4)=675241.7175watt[/tex]

The net radiant power generated by the wood-burning stove can be determined by using the Stefan-Boltzmann law. This law combines the factors of Stefan-Boltzmann constant, emissivity of the body, surface area, and the difference of the fourth powers of the absolute temperatures.

The rate of radiant power generated by the stove can be found using the Stefan-Boltzmann law, which describes the rate of heat transfer by emitted radiation. This law can be stated as P = σeAT⁴, where P is the radiant power, σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ J/s.m².K⁴), e is the emissivity of the body, A is the surface area, and T is the absolute temperature in Kelvins.

For the net rate of heat transfer by radiation (or the net radiant power) for the wood-burning stove, you would use the Stefan-Boltzmann law as Q-net = σeA(T₂⁴ - T₁⁴), where T₂ is the absolute temperature of the stove and T₁ the absolute temperature of the room.

Plugging in the stove's values from the problem: emissivity = 0.900, surface area = 3.75 m², T₂ = 455 K (stove temperature), and T₁ = 295 K (room temperature), we can calculate the net radiant power allocated by the stove.

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Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

      F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

      F = 2 6.67 10⁻¹¹ 135  5.98 10²⁴ 13 / (6.79 10⁶)³

      F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

[tex]E_g = E_p[/tex]

[tex]mgh = 0.5kx^2[/tex]

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

[tex]4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2[/tex]

[tex]47.088x + 227.906 = 1200x^2[/tex]

[tex]1200x^2 - 47.088x - 227.906 = 0[/tex]

[tex]x = 0.456m[/tex] or 45.6 cm

Answer:

∴The air cannot be made to flow in with the given pump at the given conditions.

Explanation:

Given:

The piston must be pushed more than the pressure inside the tyre:

[tex]P_g=\rho\times V\div a[/tex]

[tex]59=4.4256\times 10^{-5}\times a\times h\div a[/tex]

[tex]h=13.33\times 10^5\ in[/tex]

∴The air cannot be made to flow in with the given pump at the given conditions.

Answer:

P-value of the test statistic is 2.499

Explanation:

given data:

size sample is 24

sample mean is 414 gm

standard deviation is 15

Null hypotheis is

[tex]H_0: \mu = 420 gm[/tex]

[tex]H_1 \mu< 420[/tex]

level of significance is 0.01

from test statics

[tex]t = \frac{\hat x - \mu}{\frac{s}{\sqrt{n}}}[/tex]

degree od freedom is  =  n -1

df = 24 -1 = 13

[tex]t = \frac{414 - 420}{\frac{15}{\sqrt{24}}}[/tex]

t = -1.959

from t table critical value of t at 0.1 significane level and 23 degree of freedom  is 2.499

Answer:

Explanation:

They are infrared waves which mean they take the form of heat.

Answer: A

Explanation:

Answer:

V = 4.67 m/s

Explanation:

given,

mass of the flying bird (M)= 340 g = 0.34 Kg

Speed of bird(u₁) = 6 m/s

mass of insect(m) = 13 g = 0.013 Kg

speed of insect(u₂) = 30 m/s

Speed of bird after eating the insect(V) =?

Using conservation of momentum

M u₁ + m u₂ = (M+m)V

[tex]V = \dfrac{M u_1+ m u_2}{(M+m)}[/tex]

[tex]V = \dfrac{0.34\times 6 - 0.013 \times 30}{0.34+0.013}[/tex]

[tex]V = \dfrac{1.65}{0.353}[/tex]

V = 4.67 m/s

the bird's speed immediately after swallowing V = 4.67 m/s

Answer:

Final angular velocity, [tex]\omega_f=0.61\ rad/s[/tex]

Explanation:

Given that,

Radius of merry- go- round, r = 3 m

Inertia of the merry- go- round, [tex]I=600\ kg-m^2[/tex]

Angular speed, [tex]\omega=0.8\ rad/s[/tex]

Mass, m = 20 kg

Let I' is the new rotational inertia of the merry- go- round. Here, the angular momentum of the system remains conserved. So,

[tex]L_f=L_o[/tex]

[tex]I_f\omega_f=I_o\omega_o[/tex]

[tex]\omega_f=(\dfrac{I}{I+mr^2})\omega_o[/tex]

[tex]\omega_f=(\dfrac{600}{600+20\times 3^2})\times 0.8[/tex]

[tex]\omega_f=0.61\ rad/s[/tex]

So, the angular velocity of the merry-go-round is 0.61 rad/s. So, the correct option is (A). Hence, this is the required solution.

Explanation:

When a stone is dropped into a pond, ripples form and move in concentric circles outward. This can be treated as a point source emitting waves. And for point source emitting waves

Intensity of the wave I = Source power Ps/Area

⇒[tex]I= \frac{Ps}{\pi r^2}[/tex]

⇒I∝1/r^2

Therefore, on increasing the distance r, the intensity of the wave decreases, this also means that the amplitude also decreases.

Answer:

The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high

Explanation:

Let

[tex]d_{o}=distance of object[tex]\\f=focal length\\d_{i}=distance of image\\I_{h}=Image height[/tex]

For diverging lens:

[tex]d_{o} = 0.50\\f = -0.20\\\frac{1}{d_{o}}+\frac{1}{-0.20}\\\frac{1}{d_{i}}=\frac{1}{-0.20}-\frac{1}{0.50}=-7\\d_{o}=-\frac{1}{7}[/tex]

Magnification = [tex]\frac{d_{i}}{d_{o}}= -\frac{1}{7}÷ 0.5 = -0.286[/tex]

Image height [tex]= -0.286 * 1.5 = -0.429 cm[/tex] (negative sign means the image is virtual, inverted.

This image is [tex]\frac{1}{7}[/tex] meter to left of the center of the diverging lens.

The converging lens is located 0.08 m to the right of the diverging lens

The distance between the image of the diverging lens and center of the converging lens = [tex]\frac{1}{7} + 0.08 = 0.229 m[/tex]

The image of the diverging lens becomes the object of the converging lens.

[tex]d_{o} = 0.223\\f = 0.17\\\frac{1}{d_{i}}=\frac{1}{0.17}-\frac{1}{0.223}=0.715\\d_{i}=0.715m to the right of the converging lens[/tex]

[tex]Magnification =\frac{d_{i}}{d_{o}} = \frac{0.715}{0.223}=3.206\\image height=3.206 * 0.429 = 1.375 cm.[/tex]

The center of mass for a rectangle with a corner cut out can be calculated using the centroid formula by considering the plate and the cut-out as separate figures and then determining the average weighted by area.

This problem concerns the calculation of the center of mass for 2-dimensional figures. The center of mass represents the point at which we could balance the object in a uniform gravitational field and is calculated by integrating the mass distributed in the object.

First, consider that area mass density σ is constant. Thus, this is equivalent to finding the centroid of the shape. The location of the centroid C (x,y) is given by:

x = (A1x1 + A2x2)/(A1 + A2)

and

y = (A1y1 + A2y2)/(A1 + A2)

where Ax and Ay represent the area and centroid coordinate of the individual figure, respectively. For this problem, A1 is the whole rectangular plate and A2 is the cutout. The centroid of a rectangle is at the midpoint of its diagonals, so the x and y coordinates for A1 are (B/2, A/2) and for A2 are (D, D/2).

Then, the x-coordinate for the center of mass becomes, x = [(AB*(B/2) - CD*(D/2)]/(AB - CD) and y-coordinate for center of mass becomes, y = [(AB*(A/2) - CD*(C/2)]/(AB - CD) Calculate them to get numerical value.

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Answer:

It is a constant when the light is traveling in a vacuum.

Explanation:

Answer:

Part a)

[tex]L = 0.68 m[/tex]

Part b)

[tex]L = 0.63 m[/tex]

Explanation:

Part a)

As we know that there is no friction in the path

So here we can use energy conservation to find the distance moved by the mass

Initial spring energy = final gravitational potential energy

so we will have

[tex]\frac{1}{2}kx^2 = mgL sin\theta[/tex]

[tex]\frac{1}{2}70(0.5)^2 = 2(9.81)(L) sin41[/tex]

[tex]8.75 = 12.87 L[/tex]

[tex]L = 0.68 m[/tex]

Part b)

Now if spring is connected to the block then again we can use energy conservation

so we will have

[tex]\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2[/tex]

so we will have

[tex]\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2[/tex]

[tex]8.75 = 6.43 + 12.87 x' + 35 x'^2[/tex]

[tex]x' = 0.13 m[/tex]

so total distance moved upwards is

[tex]L = 0.5 + 0.13 = 0.63 m[/tex]

If the spring is not attached, the mass will move 0.68 meters. however, if the spring is attached, the mass will move by 0.63 meters.

[tex]\frac{1}{2} kx^2=m*g*L*sin41[/tex]

Within this formula, the symbol "k" represents 70 N/m. The "x" represents 0.50m. The "m" is 2.0 kg, the "g" represents gravity and the L is the value we need to find.

Therefore, we can substitute the values of the formula as follows:

[tex]\frac{1}{2}70(0.50)^2= 2*9.81 *L*sin41\\8.75=12.87L\\L= \frac{8.75}{12.87} = 0.67 ------- 0.68m[/tex]

[tex]\frac{1}{2} kx^2= m*g*(x+x')*sin41+\frac{1}{2} 70x'^2[/tex]

By substituting the values, we can answer the equation as follows:

[tex]\frac{1}{2} 70*0.50^2=2*9.81*(0.50+x')*sin41+\frac{1}{2} 70x'^2\\8.75=6.43+12.87x'+35x'\\X'=0.13m \\\\L=0.5+0.13=0.63 m[/tex]

More information about gravity in the link:

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Answer:

The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air

Explanation:

Due the astronaut throws the 10-kg tool kit away with a speed of 8 m/s, it gives a momentum equivalent but in the other direction, so [tex]I=mv=(10Kg)(8m/s)=80kg*m/s[/tex], then we can find the speed that the astronaut reaches due to its weight we get, [tex]v=\frac{I}{m} =\frac{80kg*m/s}{80Kg} =1m/s[/tex].

Finally, as the distance to the space shuttle is 200m, the time taken to the astronaut to reach it at the given speed will be [tex]t=\frac{d}{v}=\frac{200m}{1m/s}=200s[/tex], as the remaining air time is 4 min or 240 seconds, The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air.

Answer:

[tex]3.60432\times 10^{26}\ kg[/tex]

Explanation:

a = Orbital radius = [tex]1.62\times 10^8\ m[/tex]

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

[tex]M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg[/tex]

From the given data the mass of Saturn is [tex]3.60432\times 10^{26}\ kg[/tex]

Final answer:

To find the mass of Saturn, we can use Newton's version of Kepler's third law. Given the orbital radius and period of Saturn's moon Mimas, we can use the equation p² = a³ to calculate the mass of Saturn. Using the provided data, the mass of Saturn is approximately 5.69 x 10^26 kg.

Explanation:

To find the mass of Saturn, we can use Newton's version of Kepler's third law. Kepler's third law states that the square of the orbital period of a planet or moon is proportional to the cube of its semimajor axis. We can use the equation p² = a³ to solve for the mass of Saturn. Given that the orbital radius of Mimas, a moon of Saturn, is 1.62 × 108 m and its orbital period is about 23.21 h, we can use this data to find the mass of Saturn.

First, convert the orbital period to seconds: 23.21 h = 23.21 x 60 x 60 s = 83,556 s. Next, plug the values into the equation p² = a³: (83,556 s)² = (1.62 × 10^8 m)³. Solving for a³, we get a³ = (83,556 s)² / (1.62 × 10^8 m)³ = 38.3 x 10^18 s²/m³.

Now, rearrange the equation to solve for the mass of Saturn: mass of Saturn = (a³ x 4π²) / (G), where G is the gravitational constant. Plug in the known values: mass of Saturn = (38.3 x 10^18 s²/m³ x 4π²) / (6.67 x 10^-11 N m²/kg²). Calculating this, we get a mass of Saturn of approximately 5.69 x 10^26 kg.

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

[tex]\frac{V}{T}=const.[/tex]

Which can be also re-written as

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where

[tex]V_1, V_2[/tex] are the initial and final volumes of the gas

[tex]T_1, T_2[/tex] are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

[tex]V_1 = 7.00\cdot 10^2 mL = 700 mL[/tex] is the initial volume

[tex]T_1=20.0^{\circ}C+273=293 K[/tex] is the initial absolute temperature

[tex]V_2[/tex] is the final volume

[tex]T_2 = 1.00\cdot 10^2 K = 100 K[/tex] is the final temperature

Solving for [tex]V_2[/tex],

[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL[/tex]

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