Nuclear energy comes from splitting atoms of __________ to generate heat.

Answer: The following statement is true about squall line thunderstorm development: These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.

An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.

Final answer:

Squall lines often form ahead of advancing fronts due to warm, humid air lifting, while behind a cold front, the air is cooler and drier.

Explanation:

The statement is true concerning squall line thunderstorm development. Squall lines typically form ahead of an advancing front due to the lifting of warm, humid air in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.

Answer:

The answer is D

Hope this helps!!

Answer: D. Performing a fair test in which only the independent variable affects the dependent variable.

Explanation:

Validity is a measure of how correct the result of the experiment is. The validity of the experiment is based upon the correctness of the research topic, and correctness and implementation of the scientific procedures.  

Performing a fair test that can be based upon use of the independent and dependent variables can yield results which will be valid and acceptable. The independent variable in an experiment is the variable which can be changed to know it's affect on the dependent variable. The dependent variable will help in determining the result of the experiment.

Answer:

The correct answer is "The rigid body can have rotational and transnational motion, as long as it's transnational and angular accelerations are equal to zero."

Explanation:

A rigid body by definition does not deform when forces act on it. In case of static equilibrium a rigid body cannot have any sort of motion while in case of dynamic equilibrium it can move but with constant velocities only thus having no acceleration weather transnational or angular.

The rigid body can move in both directions as long as its angular acceleration and transverse accelerations dosent get vanished"

The phrase "angular acceleration" refers to the temporal rate at which angular velocity changes. Consequently, = d d rotational acceleration is another name for angular acceleration.

When a body experiences linear acceleration, the force or acceleration acts simultaneously on the entire body. a straight course's rate of change in velocity per unit of time. This acceleration is linear. The rotational acceleration of an item about an axis is known as angular acceleration.

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Answer:

Increased.

Explanation:

Given that in 200 mL of the beaker contains saturated solution of KI.

Whenever we allow the solution to heat or to evaporate, only water present in the solution evaporates because it is volatile while the salt is non volatile and thus do not evaporate.

When we decrease the volume of water in the beaker to 100 mL, the concentration of the salt, KI in the solution increases as the same amount of the salt is present in less amount of volume of the solvent.

[tex]f_{0}=159.2KHz[/tex]

In order to solve this problem we have to use the resonance frecuency equation [tex]f_{0}=\frac{1}{2\pi \sqrt{LC}}[/tex], at this frecuency in a LC circuit the capacitor and the inductor have the same reactance. So:

With C = 1.0μF and L = 1.0μH

[tex]f_{0}=\frac{1}{2\pi \sqrt{(1.0x10^{-6}F)(1.0x10^{-6}H)}}\\f_{0}=159155Hz\\f_{0}=159.2KHz[/tex]

The 1.0 μf capacitor and a 1.0 μh inductor have the same reactance at [tex]f_{0}[/tex] = 159.15 KHz.

A capacitor is a device that stores electrical energy.

An inductor is a device that stores electrical energy in the form of magnetic field.

Now, putting value of capacitor C and inductor L in above equation we get,

                                 [tex]f_{0}[/tex] = [tex]\frac{1}{2\pi \sqrt{LC} }[/tex]  

                                 [tex]f_{0}[/tex] = [tex]\frac{1}{2\pi \sqrt{(1.0x10^{-6})(1.0x10^{-6}) } }[/tex]

                                 [tex]f_{0}[/tex] = 159155 Hz.

Thus, for same reactance the frequency will be 159.2 KHz.

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The angular momentum quantum number can be any integer from 0 to n-1, describing the shape of the electron's orbital and the level of its angular momentum.

The angular momentum quantum number (l) can have any integer value from 0 to n-1, where n is the principal quantum number. This quantum number describes the shape of the electron's orbital, and it essentially tells us about the amount of angular momentum a subshell has. For instance, if the principal quantum number (n) is 3, l can be 0, 1, or 2, corresponding respectively to the s, p, and d orbitals.

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Answers: (A)[tex]F=G\frac{M^2}{4R^2}[/tex] (B) [tex]V=\sqrt{\frac{GM}{4R}}[/tex] (C)[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex] (D)

[tex]E=-\frac{GM^{2}}{4R}[/tex]

Explanation:

According to the law of universal gravitation:

[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]   (1)

Where:

[tex]F[/tex] is the module of the gravitational force exerted between both bodies  

[tex]G[/tex] is the universal gravitation constant.

[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.

[tex]r[/tex] is the distance between both bodies

In the case of this binary system with two stars with the same mass [tex]M[/tex] and separated each other by a distance [tex]2R[/tex], the gravitational force is:

[tex]F=G\frac{(M)(M)}{(2R)^2}[/tex]   (2)

[tex]F=G\frac{M^2}{4R^2}[/tex]   (3) This is the gravitational force between the two stars.

Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed [tex]V[/tex] of each star is the same. In addition, if we assume this system is in equilibrium, gravitational force must be equal to the centripetal force  [tex]F_{C}[/tex] (remembering we are talking about a circular orbit):

So: [tex]F=F_{C}[/tex]   (4)

Where [tex]F_{C}=Ma_{C}[/tex]  (5) Being [tex]a_{C}[/tex] the centripetal acceleration

On the other hand, we know there is a relation between [tex]a_{C}[/tex] and the velocity [tex]V[/tex]:

[tex]a_{C}=\frac{V^{2}}{R}[/tex]  (6)

Substituting (6) in (5):

[tex]F_{C}=M\frac{V^{2}}{R}[/tex] (7)

Substituting (3) and (7) in (4):

[tex]G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}[/tex]   (8)

Finding [tex]V[/tex]:

[tex]V=\sqrt{\frac{GM}{4R}}[/tex] (9) This is the orbital speed of each star

The period [tex]T[/tex] of each star is given by:

[tex]T=\frac{2\pi R}{V}[/tex]  (10)

Substituting (9) in (10):

[tex]T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}[/tex]  (11)

Solving and simplifying:

[tex]T=4\pi R\sqrt{\frac{R}{GM}}[/tex]  (12) This is the orbital period of each star.

The gravitational potential energy [tex]U_{g}[/tex] is given by:

[tex]U_{g}=-\frac{Gm_{1}m_{2}}{r}[/tex]  (13)

Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy [tex]\Delta U_{g}[/tex] for this case is:

[tex]\Delta U_{g}=U-U_{\infty}[/tex] (14)

Knowing [tex]U_{\infty}=0[/tex] the total potential energy is [tex]U[/tex] and in the case of this binary system is:

[tex]U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}[/tex]  (15)

Now, we already have the potential energy, but we need to know the kinetic energy [tex]K[/tex] in order to obtain the total Mechanical Energy [tex]E[/tex] required to separate the two stars to infinity.

In this sense:

[tex]E=U+K[/tex] (16)

Where the kinetic energy of both stars is:

[tex]K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}[/tex] (17)

Substituting the value of [tex]V[/tex] found in (9):

[tex]K=M(\sqrt{\frac{GM}{4R}})^{2}[/tex] (17)

[tex]K=\frac{1}{4}\frac{GM^{2}}{R}[/tex] (18)

Substituting (15) and (18) in (16):

[tex]E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}[/tex] (19)

[tex]E=-\frac{GM^{2}}{4R}[/tex] (20) This is the energy required to separate the two stars to infinity.

F = [tex]\frac{GM1M2}{R} \\\\[/tex]

Where F = Gravitational force that is between the two masses

M1 = mass of the first star

M2 = Mass of the second star

d= distance between the masses

G = Gravitational constant

These stars are identical so M1 = M2 = M

d = 2R

d² = 2R² = 4R²

[tex]F= \frac{GM1M2}{4R^2}[/tex]

Given that theses circles are said to move in a circular orbit, the net centripetal force

Fnet = F second law of Newton

[tex]Fnet = \frac{MV^2}{R} =\frac{GM^2}{4R^2} \\\\v^2 = \frac{GM}{4R}[/tex]

[tex]v = \sqrt[]{}\frac{{GM} }{4R}[/tex]

C. The orbital period

Period = time T

speed = distance traveled / time

distance = circumference of a circle = 2πR

V = 2πR/T

T = 2πR/V

[tex]T = 2\pi R(\sqrt{(MG)/(4R)}^{-1} \\[/tex]

[tex]T = (2\pi R\sqrt{4R} )/(\sqrt{MG} )[/tex]

[tex]T = (2\pi R(2R^{1/2} ))/\sqrt{MG}[/tex]

[tex]T = (4\pi R^{3/2}) /\sqrt{MG[/tex]

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Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.

Thermal expansion for brass = 19×10⁻⁶ °C⁻¹

Thus,

[tex]\alpha=\frac {\Delta r}{r\times \Delta T}[/tex]

Also,

[tex]\Delta T=T_{final}-T_{Initial}[/tex]

So,

[tex]19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}[/tex]

Solving for final temperature as:

[tex](T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}[/tex]

Final temperature = 106.165 °C

Answer:

1. 3.8 m/s²

2. 2400 m/s

Explanation:

1. Acceleration due to gravity can be found with Newton's law of gravitation:

g = GM / R²

where g is the acceleration at the surface,

G is the universal gravitational constant,

M is the mass of the planet,

and R is the radius of the planet.

If we say M is the mass of the Earth and R is the radius of the Earth, then we know:

9.8 = GM / R²

Mars has a mass of 0.11 M and a radius of (4200/7900) R = 0.53 R.  So the acceleration is:

g = G (0.11 M) / (0.53 R)²

g = 0.39 GM / R²

g = 0.39 (9.8)

g = 3.8

So the value of g on Mars is 3.8 m/s².

2. The escape velocity is the velocity at which the kinetic energy of an object at the surface equals its gravitational potential energy.

½ m v² = m (GM / r²) r

½ v² = GM / r

v² = 2GM / r

v = √(2GM / r)

The mass of the moon is 7.35×10²² kg, and the radius of the moon is 1.74×10⁶ m.

v = √(2 (6.67×10⁻¹¹) (7.35×10²²) / (1.74×10⁶))

v = 2400 m/s

Answer: yes that is true

Explanation:

Answer:

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is inversely proportional to the square of the distance from the light source, the form of the function is:

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

[tex] F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4[/tex]

Then, you have found that the light is 4 times as bright as it originally was.

The light becomes four times brighter when you move halfway closer to a light source, due to the inverse square law.

When you move halfway closer to a light source, the illumination on the book you are reading increases dramatically because of the inverse square law for light propagation. According to this law, the brightness or intensity of light is inversely proportional to the square of the distance from the light source. If you reduce the distance to half, the new brightness becomes (1/0.5)^2 = 4 times the initial brightness.

For example, if the initial intensity of the light is I at a distance d, when you move to a distance of d/2, the intensity becomes 4I. This is because the new distance squared is (d/2)^2 which is 1/4th of the original distance squared (d^2), hence reversing the effect we get four times the initial intensity due to the inverse law.

Answer:

A) 100°C

B) 211 g

Explanation:

Heat released by red hot iron to cool to 100°C = 130 x .45 x 645 [ specific heat of iron is .45 J /g/K]

= 37732.5 J

heat required by water to heat up to 100 °C = 85 x 4.2 x 80 = 28560 J

As this heat is less than the heat supplied by iron so equilibrium temperature will be 100 ° C. Let m g of water is vaporized in the process . Heat required for vaporization = m x 540x4.2  = 2268m J

Heat required to warm the water of 85 g to 100 °C = 85X4.2 X 80 = 28560 J

heat lost = heat gained

37732.5 = 28560 + 2268m

m = 4 g.

So  4 g of water will be vaporized and remaining 81 g of water and 130 g of iron that is total of 211 g will be in the cup . final temp of water will be 100 °C.

The final temperature of the water is 100°C

The final mass of the iron and the remaining water is 211 g

The heat released by red hot iron to cool to 100°C

= 130 x .45 x 645 [ specific heat of iron is .45 J /g/K]

= 37732.5 J

The heat required by water to heat up to 100 °C

= 85 x 4.2 x 80

= 28560 J

As this heat is less than the heat supplied by iron, the equilibrium temperature will be 100 ° C. Let mg of water that is vaporized in the process is

Heat required for vaporization

= m x 540x4.2  

= 2268m J

The heat required to warm the water of 85 g to 100 °C

= 85X4.2 X 80

= 28560 J

heat lost = heat gained

37732.5

= 28560 + 2268m

m = 4 g.

So, 4 g of water will be vaporized, and the remaining 81 g of water and 130 g of iron which is a total of 211 g will be in the cup .

The final temp of the water will be 100 °C.

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Answer:0.0242 Nm

Explanation: The distance of 0.145 mm is very small. This will also result in a small ttorque. The question would make more sense if the distance was 0.145 m. Answering with the 0.145 mm as distance :

Changin from mm to m : 1mm=0.001m

0.145 mm = 0.000145 m

The formula for Torque is T=Fxd

Wher F is the force and d is the perpendicular distance between the centre of the bolt and the applied force.

T= 167 x 0.000145 = 0.024215 Nm

Answer:

Student b is suspended at rest above the floor.

Explanation:

Student a and student b are holding the two ends of the rope which passes over a pulley.

The mass of student a = 70 kg

The mass of student b = 60 kg

The tension force acting on the rope is uniform through out the length in this case.

Force due to gravitation:

F = mg

Where,

m is the mass and g is the acceleration due to gravity

Since g is constant for both the students. Thus, F is directly proportional to the mass of the student.

Since the mass of student a is greater than student b. Thus, the force due to the weight of the student a is greater than student b.

Thus, Student a will be at the floor and Student b is suspended at rest above the floor.

Heat flows from the hot to cold substance.

Answer: Option A

Explanation:

The medium or substance which has the higher temperature referred to as the hot substance transmits heat to the less temperature body or substance referred to as the cold substance. The temperature difference makes the heat flow from warmer substance to cooler substance.

The temperature thus flows through the gradient from high to low. This transmission takes place until both the substance in contact becomes equal in temperature.

Answer:

D) multiply the force in the direction of motion by the distance the object moved

Explanation:

To calculate work done on an object, multiply the force in the direction of motion by the distance the object moved.

Final answer:

To calculate the work done on an object, multiply the force in the direction of motion by the distance the object moved, using the formula W = F d.

Explanation:

To calculate work done on an object, you multiply the force in the direction of motion by the distance the object moved. This calculation is represented by the formula W = F d, where W is the work done on the system, F is the constant force applied parallel to the direction of motion, and d is the displacement of the object in the direction of the force. It's important to note that the force considered in this calculation is only the component that acts in the same direction as the displacement of the object.

When you take US appliances to Europe, or any of the many other places in the world where the 'mains' outlets supply 240 volts, you can't just plug your device into the 240 volts.  It'll overheat, smoke, and die if you do.  

You need some kind of a power "converter", that will change the 240V to the 120V that your device recognizes from back home, and is designed to work with.

If your device uses more than just a few watts of power to operate, then you'll need a converter that's based on a transformer.  It'll be heavy and pretty bulky, but that's what it takes to do the job.

The transformer is a "step-down" unit. You put 240 volts into it, it steps that down, and 120 volts comes out of it.  

To use a hair blower designed for a 120V supply in a region with a 240V supply, you need a step-down voltage converter, which reduces the voltage to a level that the device can handle.

To operate a hair blower designed for 120V in a region which provides 240V, you need to use a device known as a voltage converter or power converter. As you want to reduce the voltage from 240V to 120V, you would need a step-down converter.

The converter works by reducing the voltage offered by the power source (in this case, 240V) to a level the device can handle (120V). Without such a converter, plugging the hair blower in directly would likely cause it to be ruined, as it isn't constructed to handle such a high voltage.

It’s also important to remember that different regions have different power plug formats, so you may also need to carry a plug adapter.

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Answer:

Coefficient of static friction is normally greater than coefficient of sliding friction which is also known as coefficient of kinetic friction.

Explanation:

Coefficient of friction can be seen as the cumulative effect of the irregularities that are on the surface of objects. When these interlocking get locked into one another a resistance arises to the motion of the object which is termed as friction. When an object is static these irregularities get more time to be interlocked as compared to when an object is in motion thus the coefficient of static friction is more than the coefficient of sliding friction.

Static friction is normally greater than sliding friction on the same object because it takes more force to start an object moving than to keep it moving.

In general, static friction is normally greater than sliding friction on the same object. Static friction occurs when two surfaces are in contact but not moving relative to each other, while sliding friction occurs when two surfaces are sliding past each other.

The force of static friction is needed to prevent the object from moving, and it can be greater than the force of sliding friction because it takes more force to start an object moving than to keep it moving. This is why it is often harder to start sliding a heavy object compared to keeping it sliding.

For example, imagine trying to push a heavy box on the ground. Initially, you need to apply a greater force to overcome the static friction and start the box moving. Once the box is sliding, the force required to keep it sliding (sliding friction) is usually less than the force needed to start it moving (static friction).

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In an isothermal expansion of a monatomic ideal gas where the volume doubles, Boyle's Law implies that the pressure of the gas will decrease by a factor of 2, or in other words, will be halved.

The change in pressure for a monatomic ideal gas in an isothermal expansion where the volume doubles can be explained using Boyle's Law. Boyle's Law states that the product of the pressure and volume for a given gas sample is constant as long as the temperature is constant (P₁V₁ = P₂V₂). Therefore, if the volume of the gas doubles, the pressure would reduce by half, assuming the temperature remains constant.

Using the ideal gas equation, p = nRT/V, where 'p' is the final pressure, 'n' is the number of moles, 'R' is the gas constant, 'T' is temperature, and 'V' is volume, we find the final pressure of the gas to be Po/2, where Po is the initial pressure.

Thus, in an isothermal expansion of an ideal monatomic gas where the volume doubles, the pressure will decrease by a factor of 2, that is, it will be halved.

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Answer:

2144 rad/s

Explanation:

R1 = R

ω1 = 536 rad/s

R2 = R/2

ω2 = ?

Mass is M

By use of angular momentum remains constant if no external force is acting on the body.

I1 ω1 = I2 ω2

The moment of inertia of solid sphere is 12/5 MR^2

So, 2/5 x M R^2 x 536 = 2/5 x M (R/2)^2 x ω2

536 = ω2 / 4

ω2 = 2144 rad/s

Answer:

ω₂ = 2144 rad/s

Explanation:

angular  speed =  536 radians/second

as, we all know the moment of inertia of solid sphere

[tex]I_{sphere}= \dfrac{2}{5}MR^2[/tex]

here in the question two radius are given

by using angular momentum conservation

[tex]I_1 \omega_1 = I_2 \omega_2[/tex]

[tex]\dfrac{2}{5}MR_1^2 \omega_1 =\dfrac{2}{5}MR_2^2 \omega_2\\R^2\times 536= \dfrac{R^2}{4}\times \omega_2[/tex]

[tex]\omega_2 = 4 \times 536[/tex]

ω₂ = 2144 rad/s

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

[tex]h_{fg}[/tex] = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

[tex]T_{L}[/tex] = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

[tex]T_{H}[/tex] = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, [tex]q_{reject}[/tex] = [tex]h_{fg}[/tex] = 171.82 kJ/kg

We know COP of heat pump

COP = [tex]\frac{T_{H}}{T_{H}-T_{L}}[/tex]

        = [tex]\frac{304.31}{304.31-254.23}[/tex]

         = 6.076

Therefore, Work out put, W = [tex]\frac{q_{reject}}{COP}[/tex]

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

Final answer:

Accurately calculating the net work input for a heat pump using the reversed Carnot cycle requires additional data beyond the pressure limits and the typical phase change of R-134a. Specific temperatures, heat transfers, or detailed thermodynamic tables for R-134a are essential for this calculation.

Explanation:

To determine the net work input for a heat pump operating on the reversed Carnot cycle with R-134a as the working fluid, it's important to understand the foundational concepts of heat pumps and thermodynamics. Heat pumps transfer heat from a colder area (cold reservoir) to a warmer area (hot reservoir). When examining a reversed Carnot cycle on a P-V diagram, the area within the cycle represents work, however, since it is reversed, the location of heat transfer and work are also reversed compared to a regular heat engine. In a standard heat engine, work is output, but in a heat pump, work is input.

The net work input is calculated by considering the efficiency of the system, which is dependent on the temperatures of the cold and hot reservoirs. However, since the question does not provide specific temperatures or heat transfers but only pressures, and asks us to consider a scenario where the working fluid, R-134a changes from saturated vapor to saturated liquid during the heat-rejection process, additional information such as thermodynamic tables for R-134a or further details about the specific amounts of heat transfer would be needed to calculate the precise net work input.

Answer: (a)[tex]F=7(10)^{-7}N[/tex]

              (b)[tex]F=1.344(10)^{-6}N[/tex]  

              (c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex]   (1)

Where:

[tex]F[/tex] is the module of the force exerted between both bodies

[tex]G[/tex] is the universal gravitation constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.

[tex]r[/tex] is the distance between both bodies

Knowing this, let's begin with the answers:

Using equation (1) and taking into account the mass of the father [tex]m_{1}=100kg[/tex], the mass of the baby [tex]m_{2}=4.20kg[/tex] and the distance between them [tex]r=0.2m[/tex], the force [tex]F_{F}[/tex]  exerted by the father is:

[tex]F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}[/tex]   (2)

[tex]F_{F}=0.0000007N=7(10)^{-7}N[/tex]   (3)

Using again equation (1) but this time taking into account the mass of Jupiter [tex]m_{J}=1.898(10)^{27}kg[/tex], the mass of the baby [tex]m_{2}=4.20kg[/tex] and the distance between Jupiter and Earth (where the baby is) [tex]r_{E}=6.29(10)^{11}m[/tex], the force [tex]F_{J}[/tex]  exerted by the Jupiter is:

[tex]F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}[/tex]   (4)

[tex]F_{J}=0.000001344N=1.344(10)^{-6}N[/tex]   (5)

Now, comparing both forces:

[tex]F_{J}=0.000001344N=1.344(10)^{-6}N[/tex]   and [tex]F_{F}=0.0000007N=7(10)^{-7}N[/tex]  we can see [tex]F_{J}[/tex] is greater than [tex]F_{F}[/tex]. However, the difference is quite small as well as the force exerted on the baby.

The gravitational force a 100 kg father exerts on a 4.20 kg baby is approximately 0.014 N, while Jupiter's gravitational force on the baby is just around 9.36 * 10^-11 N. This implies that the father's gravitational force on the baby is significantly greater than that of Jupiter.

This question involves calculating the gravitational force, which according to Newton's Law of Gravitation is given by the formula F = G*(m1*m2)/r^2, where G is the universal gravitational constant (6.674 * 10^-11 N(m/kg)^2), m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

(a) For the father and the baby, substitute in the given values into the formula, F (father) = (6.674 * 10^-11)*(4.20*100)/0.200^2 = 0.014 N.

(b) For Jupiter and the baby, know that the mass of Jupiter is about 1.898 × 10^27 kg, F (Jupiter) = (6.674 * 10^-11)*(4.20*1.898 × 10^27)/(6.29 * 10^11)^2 which is approximately 9.36 * 10^-11 N.

(c) The comparison of the forces shows that the gravitational force the father exerts on the baby is considerably larger than the gravitational force Jupiter exerts on the baby, reinforcing the observation that gravity is a relatively weak force that is only noticeable when dealing with massive objects.

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Answer:

option (c)

Explanation:

The time taken by the pendulum to complete one oscillation is called time period.

The number of oscillations completed in one second is called frequency.

The frequency is the reciprocal of time period.

T = 1 / f

If the period is doubled, then the frequency is halved.

Final answer:

To find the initial temperature, we can use the combined gas law. By plugging in the values of the initial and final volumes and the final temperature, we can solve for the initial temperature.

Explanation:

To solve this problem, we can use the combined gas law, which states that for an ideal gas at constant pressure, the ratio of the initial volume and the initial temperature is equal to the ratio of the final volume and the final temperature:

V1/T1 = V2/T2

Given that the initial volume is 2.70 L, the final volume is 1.75 L, and the final temperature is 17.00 °C, we can plug these values into the combined gas law to find the initial temperature:

2.70/T1 = 1.75/17.00

Solving for T1, we get T1 = 45.545 °C. Therefore, the initial temperature of the gas was approximately 45.545 °C.

Answer:

[tex]\omega_f=0.0356 rev/day[/tex]

Explanation:

Given:

Angular speed [tex]\omega_1=2\ rev/day[/tex]

Radius of the solid sphere = R

Radius of the spherical shell, R' = 5.8R

now the initial moment of inertia i.e the moment of inertia of the solid sphere is given as:

[tex]I_i=\frac{2}{5}MR^2[/tex]

where, M is the mass of the solid sphere

Now, the final moment of inertia i.e the moment of inertia of the spherical shell is given as:

[tex]I_f=\frac{2}{3}MR'^2[/tex]

or

[tex]I_f=\frac{2}{3}M(5.8R)^2[/tex]

or

[tex]I_f=22.426MR^2[/tex]

Now applying the concept of conservation of angular momentum

we get

[tex]I_i\omega_i=I_f\omega_f[/tex]

substituting the values, we get

[tex]\frac{2}{5}MR^2\times 2=22.426\times MR^2\omega_f[/tex]

or

[tex]\omega_f=\frac{\frac{2}{5}\times 2}{22.426}=0.0356 rev/day[/tex]

Final answer:

To find the mass of the baseball, we can use the concept of impulse, which is equal to the change in momentum. We are given the net eastward impulse delivered by the bat and the initial and final velocities of the baseball. By using the equation for impulse, we can calculate the mass of the baseball to be 170 grams.

Explanation:

To find the mass of the baseball, we can use the concept of impulse. Impulse is equal to the change in momentum, which is the product of mass and velocity. In this case, we are given the net eastward impulse delivered by the bat (1.5 N-s) and the initial and final velocities of the baseball (-3.8 m/s and 4.9 m/s, respectively).

Since the impulse is equal to the change in momentum, we can write the equation:

Impulse = (mass of the baseball)(final velocity - initial velocity)

Substituting the given values, we have:

1.5 = (mass of the baseball)(4.9 - (-3.8))

Simplifying the equation and solving for the mass of the baseball:

mass of the baseball = 1.5 / (4.9 - (-3.8))

mass of the baseball = 1.5 / 8.7

mass of the baseball = 0.17 kg

However, the question asks for the mass of the baseball in grams. Therefore, we need to convert the mass from kilograms to grams:

mass of the baseball = 0.17 kg * 1000 g/kg

mass of the baseball = 170 g

Answer:

109.09°C

Explanation:

Given data:

The capacity of the car cooling system =  4.40 gallons

Ratio of water and solute = 50/50

thus,

Volume of solute = Volume of water = 4.40/ 2 gallons = 2.20 gallons

thus, the Mass of solute, M₁ = Density × Volume = 1.1 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 9160.69 grams

Mass of water = 0.998 g/mL × [tex]2.20\times\frac{3785.41\ \textup{mL}}{1\ \textup{gallon}}[/tex] = 8310.346 grams

molality , m = [tex]\frac{mass\ of \ solute\ \times 1000}{molar\ mass\ of\ solute\times\ mass\ of\ water}[/tex]

on substituting the values, we get

m = [tex]\frac{9160.69 \times 1000}{62.07\times8310.346}[/tex] =  17.759 m

Now, the elevation in boiling point is given as:

[tex]\Delta T_b = k_bm[/tex]

where,

[tex]k_b[/tex] is a constant = 0.512 °C/m

on substituting the values we get

Boiling point = 100°C + 17.759 × 0.512= 109.09°C

The boiling point of the solution will be equal to 109.09°C.

We can arrive at this answer as follows:

First, we must take into account that the water and solute ratio is 50/50, therefore, we must divide the capacity of the car's cooling system by 2, to know the volume of solute and solvent, which will be equal per account of the reason between them.

[tex]\frac{4.40}{2} = 2.20[/tex] gallons

[tex]Density * Volume\\1.1*(2.20*3785.41)\\9160.69 g[/tex]

[tex]0.998*(2.20*3585.41)\\8310.346 g[/tex]

[tex]Molality = \frac{M_s*1000}{molar M_S*M_W}\\\\Molality = \frac{9160.69*1000}{62.07*8310.34} \\\\Molality= 17.759 m[/tex]

[tex]\Delta T= K_b*molality\\[/tex]

We must take into account that [tex]K_b[/tex] is a constant and its value is always 0.512 °C/m.

[tex]\Delta T= 0.512*(100C + 17.759)\\\Delta T= 109.09C[/tex]

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Answer:

There is no way by which we can tell weather we are on the earth or we left the earth.

Explanation:

Our body responses to acceleration alone when we accelerate upwards at [tex]9.8m/s^{2}[/tex] a fictitious force equal to our [tex]mass\times acceleration[/tex] acts on our body which in this case of acceleration is same as our weight thus we will still feel the effect of gravity thus cannot say weather we left earth or are still there.

The energy required to raise the temperature of Lake Erie's water from 17.8°C to 21.6°C is 1.96×10¹⁶ J. It would take approximately 4.11×10^9 years to supply this amount of energy using a 1,400-MW exhaust energy of an electric power plant.

To calculate the energy required to raise the temperature of Lake Erie's water from 17.8°C to 21.6°C, we need to use the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given that Lake Erie contains 4.00 ✕ 10^11 m³ of water and assuming the density of water at 20°C and 1 atm to calculate mass, we can find the energy. Using the formula, Q = mcΔT, we can substitute the values and calculate the energy as 1.96×10¹⁶ J.

For part (b), we need to determine how many years it would take to supply this amount of energy using a 1,400-MW exhaust energy of an electric power plant. The formula to calculate the energy supplied by the power plant over a certain time period is E = Pt, where E is the energy, P is the power, and t is the time. Rearranging the formula to solve for time, t = E/P, we can substitute the values and calculate the time as approximately 4.11×10^9 years.

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The student's question requires us to use the equation for heat change to calculate the energy needed to raise the temperature of Lake Erie by a few degrees. We also need to calculate the time it would take for a power plant to produce this amount of energy. The answers we found are approximately 6.36 × 10^24 joules and 1.44 × 10^8 years respectively.

To answer the student's question, we first need to calculate the energy required to raise the temperature of Lake Erie from 17.8°C to 21.6°C. The formula for this calculation involves the specific heat of water, the mass of the water, and the change in temperature. Specifically:

ΔQ = m × c × ΔT

where m is the mass of the water, c is the specific heat of water (4.184 J/g°C), and ΔT is the temperature change (21.6°C - 17.8°C = 3.8°C). Since the volume of Lake Erie is given as 4.00 × 10^11 m³ and we're assuming the density of water is about 1 g/cm³, we multiply the volume by the density to get approximately 4.00 × 10^20 g. Our calculation then becomes:

ΔQ = (4.00 × 10^20 g) × (4.184 J/g°C) × (3.8°C) = approximately 6.36 × 10^24 J.

To answer the second part of the question, we need to convert the power of the power plant from MW to J/s. 1 MW is equivalent to 1 × 10^6 J/s, so the plant is producing 1.4 × 10^9 J/s. To find out how long it would take to produce the energy calculated in the first part, we simply divide the total energy requirement by the energy production rate of the power plant, which gives us:

Time = (6.36 × 10^24 J) / (1.4 × 10^9 J/s) = approximately 4.54 × 10^15 seconds. Converting seconds to years, we get approximately 1.44 × 10^8 years.

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The force applied perpendicular to the door exerts more torque, because all of it is used for rotation. When a force is applied at an angle, only the part of the force acting perpendicular to the door contributes to the torque.

The force that exerts more torque on the door in this case is the one applied perpendicular to the door. This is because torque is calculated as the product of the force applied and the distance from the pivot point at which it is applied. The key point here is that only the component of the force acting perpendicular to the door creates torque. When a force is applied at an angle, only the component of the force acting perpendicular to the door will create a torque, hence reducing the total torque. So, in this case, the first force applied perpendicular to the door exerts greater torque because the full magnitude of the force is acting to rotate the door.

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