At what fraction of its current radius would the free-fall acceleration at the surface be three times its present value?

Answer:

2.5 seconds

Explanation:

s(t) = -16t^2 + 80t + 384

for

0≤t≤8

First we differentiate s(t) to get s'(t)

s'(t) = -32t + 80

Let us then find the critical point; thus we will equate s'(t) to zero and then search for values where s'(t) is undefined

s'(t) = -32t + 80 = 0

t = 80/32

t = 2.5 sec

Let us evaluate s at the critical points and end points

s(0) = -16(0)^2 + 80(0) + 384 = 384

s(2.5) = -16(2.5)^2 + 80(2.5) + 384 = 684

s(8) = -16(8)^2 + 80(8) + 384 = 0

Thus, the stone attains it maximum height of 684ft at at t=2.5s

Answer: answer is option A

Explanation: gradually model ascertained the features of the surface of mass is explaining it as gradual incremental changes continuously over a long period of time which negates the catastrophic model nature of turtles model

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]

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.

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:[tex]5 rad/s^2[/tex]

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

[tex]Torque =I\alpha =F\cdot r[/tex]

where [tex]\alpha =angular\ acceleration[/tex]

[tex]I\alpha =F\cdot r[/tex]

[tex]0.02\cdot \alpha =1\cdot 0.1[/tex]

[tex]\alpha =5 rad/s^2[/tex]    

The angular acceleration of the cylinder is 5 rad/s².

To calculate the angular acceleration of the cylinder, we use the formula of torque below.

Torque is the force that causes a body to rotate about an axis.

Formula:

Where:

Make α the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The angular acceleration of the cylinder is 5 rad/s².

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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:

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

Explanation:

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]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]

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Answer:

The time taken to rotate the sphere one time is,  t = 22 s

Explanation:

Given data,

The mass of the sphere, m = 8200 kg

The radius of the sphere, r = 90 cm

                                             = .9 m

The force applied by the girl, F = 75 N

The moment of inertia of the sphere is,

                            I = 2/5 mr²

                              = (2/5) 8200 x (.9)²

                              = 2657 kg·m²

The torque,

                            τ = I α

                             75 x 0.9 = 2657 x α

                              α = 0.0254 rad/s²

The angular displacement,

                            θ = ½αt²

                             2π =  ½ x 0.0254 rad/s² x t²

                                t = 22 s

Hence, the time taken to rotate the sphere one time is,  t = 22 s

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:

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:

Evaporation is the physical change in which a substance converts from its liquid state to its gaseous state. Condensation is the physical change in which a substance converts from its gaseous state to its liquid state.

Explanation:

Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.

Answer:

it takes the sun about 230 million years to orbit the milky way.

Explanation:

we're moving at an average velocity of 828,000 km/hr but it still takes us a long time to orbit the milky way.

Answer:

[tex]SE_{gel} = 3.75[/tex]

Explanation:

First, we have to calculate the gel's column height using the cylinder's volume, as follows:

[tex]V=\pi\times r \times h\\h=\frac{V}{\pi \times r}\\h=\frac{4.1m^3}{\pi \times 1m}= 1.30 m[/tex]

Then, as the pressure given at the bottom of the tank is the sum of the surface pressure and the gel's column pressure, we need to calculate only the gel's column pressure:

ft of water is a unit of pressure, but we need to convert it to atm and then to Pa, in order to calculate our results in the correct units. Therefore, the conversion factor is:

1 ft of water (4°C) = 0.0295 atm

[tex]60 ft water \times \frac{0.0295 atm}{1 ft water}= 1.77  atm\\P_{bottom}=P_{surface}+P_{gel}\\P_{gel}=P_{bottom}-P_{surface}=1.77 atm - 1.3 atm\\P_{gel}= 0.47 atm\times \frac{101325Pa}{1 atm}=47622.75 Pa[/tex]

Now, to calculate the specific gravity, we need to find first the gel's density:

[tex]P_{gel} = \rho gh\\\rho = \frac{P_{gel}}{gh}=\frac{47622.75 Pa}{9.8 m/s^2 \times 1.30m}= 3738.04 \frac{kg}{m^3}[/tex]

[tex]SE_{gel} = \frac{\rho_{gel}}{\rho_{water}}= \frac{3738.04 kg/m^3}{997 kg/m^3} = 3.75\\SE_{gel} =3.75[/tex]

The specific gravity of the gel is 3.75.

Final answer:

To calculate the specific gravity of a gel, convert the pressure reading from feet of water to pascals, add the tank's pressurized atmosphere, then use the relation between pressure, density, gravity, and height to solve for the density of the gel, which is then compared to the density of water.

Explanation:

The question is asking to calculate the specific gravity of a gel based on the pressure reading from a probe at the bottom of a cylindrical tank. We are given the pressure as 60 ft of water. To convert this to a pressure that we can use to find density, we need to convert feet of water to pascals:

1 ft H2O = 2989.07 Pa

60 ft H2O = 60 * 2989.07 Pa = 179344.2 Pa

Since the tank is pressurized to 1.3 atmospheres at the surface, we must also consider this in our pressure calculation:

1 atm = 101325 Pa

1.3 atm = 1.3 * 101325 Pa = 131722.5 Pa

The total pressure at the bottom of the tank is the sum of the pressure due to the gel and the pressurized air:

Total pressure = 179344.2 Pa + 131722.5 Pa

Total pressure = 311066.7 Pa

To find the specific gravity, we use the following relation where the specific gravity is the ratio of the density of the gel (ρgel) to the density of water (ρH2O):

Specific gravity = ρgel / ρH2O

We know that pressure is also the product of density (ρ), gravity (g=9.81 m/s²), and height (h=60 ft * 0.3048 m/ft) for the fluid:

Pressure = ρgel * g * h

ρgel = Pressure / (g * h)

ρgel = 311066.7 Pa / (9.81 m/s² * 60 ft * 0.3048 m/ft)

After calculating ρgel, we divide that by the density of water (1000 kg/m³) to find the specific gravity.

Answer:

Explanation:

Given

Each student exert a force of [tex]F=400 N[/tex]

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students [tex]F=18\times 400=7200 N[/tex]

therefore weight of car [tex]W=7200[/tex]

mass of car [tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{7200}{9.8}=734.69 kg[/tex]

(b)[tex]7200 N \approx 1618.624\ Pound-force[/tex]

[tex]734.69 kg\approx 1619.71 Pounds[/tex]                  

The mass of the car lifted by eighteen students is 734.69 kg, and its weight is approximately 1620.29 pounds.

To determine the mass of the car that eighteen students lift in a college homecoming competition, given that each student exerts an upward force of 400 N, first calculate the total force exerted by the students: 18 students × 400 N per student = 7200 N.

Next, we use the equation for weight (W = mg) to find the mass (m), where W is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s²). Thus, the mass of the car is m = W/g = 7200 N / 9.8 m/s² ≈ 734.69 kg.

To find the weight in pounds, we can use the conversion factor 1 kg ≈ 2.20462 lbs. Therefore, the weight of the car in pounds is approximately 734.69 kg × 2.20462 lbs/kg ≈ 1620.29 lbs.

Answer:

[/tex] 408.8[/tex]

Explanation:

[tex]t[/tex] = thickness of the floor = 0.159 m

[tex]k[/tex] = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹

[tex]T_{i}[/tex] = Temperature of the interior of the cabin = 18.0 ⁰C

[tex]T_{o}[/tex] = Temperature of the air below the cabin = - 16.4 ⁰C

Difference in temperature is given as

[tex]\Delta T[/tex] = Difference in temperature = [tex]T_{i} - T_{o} = 18 - (- 16.4) = 34.4[/tex]⁰C

[tex]A[/tex] = Area of the floor = 13.4 m²

[tex]Q[/tex] = Rate of heat conduction

Rate of heat conduction is given as

[tex]Q = \frac{kA \Delta T}{t}[/tex]

[tex]Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8[/tex] W

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:

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:

Answer:

 y = -19.2 sin (23.15t) cm

Explanation:

The spring mass system is an oscillatory movement that is described by the equation

      y = yo cos (wt + φ)

Let's look for the terms of this equation the amplitude I

     y₀ = 19.2 cm

Angular velocity is

     w = √ (k / m)

     w = √ (245 / 0.457

     w = 23.15 rad / s

The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

The speed of the equation is obtained by the derivative with respect to time

     v = dy / dt

     v = - y₀ w sin (wt + φ)

For t = 0

     -vo = -yo w sin φ

The angular and linear velocity are related v = w r

      v₀ = w r₀

      v₀ = v₀ sinφ

      sinφ = 1

      φ = sin⁻¹ 1

      φ = π / 4    rad

Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

       y = -19.2 sin (23.15t) cm

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:

Current divider

Explanation:

To calculate current flow through any branch of a circuit by substituting the values of [tex]I_X\ and\ R_X[/tex] for the branch  values when total circuit current and the resistance are known, use the current divider formula.

A current divider is a circuit that produces output current as a function of input current. It is a rule to find the splitting of the current in all branches of the circuit. Hence, the correct option is (B).  

The formula to calculate the current flow through any branch of a circuit when the total circuit current and resistance are known is the current divider formula.

To calculate the current flow through any branch of a circuit by substituting the values of IX and RX for the branch values when total circuit current and the resistance are known, use the current divider formula. The current divider rule is particularly useful in parallel resistor circuits and it allows for the easy calculation of the current flowing through a resistor in parallel.

In a parallel circuit, the voltage across each branch is the same, but the current through each branch can be different, depending on the resistance of that branch. The current divider rule states that the current through a branch is the ratio of the total parallel resistance to the branch resistance, times the total current entering the parallel combination.

For example, if we want the current through resistor R1 (I1), and we know the total current (Itot) and the total parallel resistance (Rp), as well as the resistance of R1 (R1), we can use the formula:

I1 = Itot * (Rp / R1)

By knowing the total resistance and the total current, one can deduce the individual branch currents using Ohm's Law and the principles of parallel circuits.

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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:

Speed of the ambulance is 32.7 metres per second.

Explanation:

Let the actual frequency of the siren be [tex]f_{0}[/tex].

Frequency observed by me when ambulance is approaching = 540 Hz

Frequency observed by me when ambulance is moving away = 446 Hz

Let [tex]v_{s}[/tex] be the speed of sound and [tex]v_{a}[/tex] be the speed of ambulance.Then according to Doppler effect:

When source is moving towards observer,frequency observed is given as

[tex]f_{0} \times \frac{v_{s} }{v_{s} - v_{a} }[/tex] = 540 Hz    

When source is moving away from observer,frequency observed is given as

[tex]f_{0} \times \frac{v_{s} }{v_{s} + v_{a} }[/tex] = 446 Hz    

Taking [tex]v_{s} = 343 \frac{m}{s}[/tex] and solving the above two equations by eliminating [tex]f_{0}[/tex],

we get [tex]v_{a} = 32.7 \frac{m}{s}[/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:

11.16 m

Explanation:

total distance traveled by the wave = speed * time

                                                            = 0.720 m/s * 31.0 s

                                                            = 22.32 m

Since this is an echo back to the starting point then the length of the swimming pool must be half of the distance traveled.

length of swimming pool  =  22.32 m÷2 = 11.16 m

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:

(1) The ____astrometric method_______ was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.

(2) Discovering planets through the ____doppler method ______ requires obtaining and studying many spectra of the same star.

(3) The____kepler mission ____________ successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.

(4) The ___transit method_________ is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.

(5) Compared to the planets of our solar system, the composition of a ____water world _____________ most resembles the compositions of Uranus and Neptune.

(6) Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call__migration ____.

(7) An extrasolar planet that is rocky and larger than Earth is often called a__super-Earth ___.

The Doppler technique and transit technique are used to discover extrasolar planets such as Jupiter-sized ones and those larger than Earth respectively. The Kepler mission has been successful in identifying thousands of these. The composition of extrasolar planets can differ, some resembling Uranus and Neptune, while the concept of 'planet migration' explains the closeness of large planets to their stars.

1. The Doppler technique was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.

2. Discovering planets through the spectroscopy requires obtaining and studying many spectra of the same star.

3. The Kepler mission successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.

4. The transit technique is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.

5. Compared to the planets of our solar system, the composition of a mini-Neptune most resembles the compositions of Uranus and Neptune.

6. Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call planet migration.

7. An extrasolar planet that is rocky and larger than Earth is often called a super-Earth.

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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

Learn more about energy:

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