A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door. Where is the pivot point and what is the maximum torque?

Answer:

f₂ = 0.019 m

Explanation:

Let's analyze this exercise a bit, when taking a picture the image should always be in the same place, position of the CCD, let's use the builder's equation to find this distance from the image (i)

         1 / f = 1 / o + 1 / i

Where f is the focal length and "o, i" are the distances to the object and image, respectively

         1 / i = 1 / f - 1 / o

Let's reduce the magnitudes to the SI system

         f = 50 mm = 0.050 m

         o = 5.0 m

Let's calculate

        1 / i = 1 / 0.050 - 1 / 5.0 = 20- 0.2 = 19.8

         i = 0.020 m

Now the object is 60 cm, rotates the lens and has a new focal length

        o₂ = 60 cm = 0.60 m

        1 / f = 1 / 0.60 + 1 / 0.020 = 1.66 + 50 = 51.66

        f₂ = 0.019 m

Final answer:

Measuring 20 cycles of the pendulum motion to determine the period instead of just one cycle allows for a more accurate and representative measurement. By averaging out any errors or variations in the measurement process, the value for the period of the pendulum is more precise. Measuring multiple cycles also helps identify any potential changes in the period over time.

Explanation:

The reason we measure 20 cycles of the pendulum motion to determine the period, rather than just one cycle, is to obtain a more accurate and representative measurement of the period. By measuring multiple cycles, we can average out any errors or variations in the measurement process, resulting in a more precise value for the period of the pendulum.

For example, if we were to measure only one cycle of the pendulum, any small errors in the timing or counting of the cycles could significantly affect our measurement. However, by measuring 20 cycles and then dividing the total time by 20, we can minimize these errors and obtain a more reliable measurement of the period.

Furthermore, measuring multiple cycles allows us to observe any potential changes in the period over time. In certain situations, the period of a pendulum may vary slightly due to factors such as air resistance or changes in the length of the string. By measuring multiple cycles, we can identify and account for any such variations, providing a more comprehensive understanding of the pendulum's behavior.

The electric flux through one of the faces of the cube with the charge at its geometric center can be found using Gauss's Law. The electric flux  enclosed by the surface divided by the permittivity of the medium is 1.424 N·m2/C.

We can use Gauss's Law to determine the electric flux through one of the faces of the cube with the charge at its geometric center. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of the medium.

The charge enclosed by one face of the cube is 12.6 µC. The permittivity of a vacuum is 8.85419 × 10-12 C2/N·m2.

Therefore, the electric flux through one face of the cube is (12.6 µC) / (8.85419 × 10-12 C2/N·m2) = 1.424 N·m2/C.

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Answer

given,

distance of fall = h

initial speed = 0 m/s

it travels 0.540 h in the last 1.00 s

Speed of the fall = [tex]\dfrac{Distance}{time}[/tex]

Speed = [tex]\dfrac{0.540\ h-0}{1}[/tex]

S = 0.540h m/s

time of the fall

using equation of motion

v = u + g t

0.540 h = 0 + 9.8 t

t = 0.055h s

The time of fall is equal to 0.055h s

height of fall = ?

again using equation of motion

v² = u² + 2 g s

(0.540h)² = 0 + 2 x 9.8 x s

s = 0.0148 h²

Hence, the time of fall of the object = 0.055h s.

            the distance of fall of object = 0.0148 h²

The object has been falling for a total time of 3.13 seconds and the total height from which it fell is 47.9 meters, given that it covered 0.540 of its total height in the last 1 second of free fall.

Let's tackle this physics problem step by step to find both the time the object has been falling and the total height from which it fell. We'll use the known acceleration due to gravity (g = 9.81 m/s²) and apply kinematic equations.

Given that the object travels 0.540h in the last 1.00 s, we can set up two equations using the kinematic formula h = ½gt², where h is the height, g is the acceleration due to gravity, and t is the time in seconds:

We solve the second equation for t, which will then be used to find the total height using the first equation.

Solving the second equation: 0.540h = ½g(t²) - ½g((t-1)²)

Expand and simplify the equation: 0.540h = ½g(t²) - ½g(t² - 2t + 1)

0.540 = t² - ½g(t² - 2t + 1)

Now, assuming g is approximately 9.8 m/s², we can plug in values and solve for t:

0.540 = t² - 0.5(9.81)(t² - 2t + 1)

After solving for t, we find:

This calculation assumes that air resistance is negligible and the acceleration due to gravity is constant.

An atom in normal conditions refers when electrons are in the fundamental state. When you leave the atom, an electron absorbs energy from an external source and moves to a higher energy state.

Energy states or energy levels are called orbitals. The difference between the energy states in an atom is responsible for the emission of photons when an electron transition occurs between these two energy states.

Because the energy levels are discrete, the emitted photons also possess different energies.

Answer:

Explanation:

Given

Mass of big box is M and small box is m

Tension T will cause the boxes to accelerate

[tex]T=(M+m)a[/tex]

where a=acceleration of the boxes

Now smaller box will slip over large box if the acceleration force will exceed the static friction

i.e. for limiting value

[tex]\mu _smg=ma[/tex]

[tex]a=\mu _s\cdot g[/tex]

thus maximum tension

[tex]T=\mu _s(M+m)g[/tex]

Answer: time(t) = 1.72s

Explanation: [tex]Angular acceleration =\frac{angular velocity}{time taken}[/tex]

Let angular acceleration = α = [tex]2.69rad/s^{2}[/tex]

ω = angular velocity = 44rev/ min

the angular velocity is in rev/ min but we need to have it in rad/s , thus we do so below

recall that 1 rev = 2π  and 1 min = 60s, [tex]\frac{44.2 * 2\pi }{60} \\\\ \frac{44.2 * 2*3.142}{60} \\\\= 4.63rads^{-1}[/tex]

hence 44.2rev/min = 44.2 * 2π/ 60 = 44.2 * 2 *3.142/ 60

thus angular velocity = 4.63rad/s

time taken = angular velocity/ angular acceleration

time taken = 4.63/2.69

time taken = 1.72s

Explanation:

Expression for energy balance is as follows.

        [tex]\Delta E_{system} = E_{in} - E_{out}[/tex]

or,          [tex]E_{in} = E_{out}[/tex]

Therefore,  

         [tex]m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})[/tex]

          [tex]h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}[/tex]

Hence, expression for exit velocity will be as follows.

           [tex]V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}[/tex]

                      = [tex]V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]

As [tex]C_{p}[/tex] for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

       [tex]V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]                    

                  = [tex][(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}[/tex]

                 = 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

[tex]\sum F = ma[/tex]

[tex]T-mg = ma[/tex]

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

[tex]T = m(g+a)[/tex]

[tex]T = (975)(9.8+0.754)[/tex]

[tex]T= 10290.15N[/tex]

Therefore the tension force in the cable is 10290.15N

Answer:

42 minutes

Explanation:

First, let us find out the time required by the roommate, who is driving the truck, to reach the city.

We know that,

[tex]time=\frac{distance}{speed} =\frac{360}{50} hrs=7.2hrs[/tex]

Now, you are planning to rest a bit after traveling for an hour. So,

distance covered in that 1 hour = [tex]60mi/h\times1h=60 miles[/tex]

time required by you to cover the total distance = [tex]\frac{360mi}{60mi/h} =6hours[/tex]

If you wish reach at the destination half an hour (0.5 h) before your roommate, you can expend a total of [tex]7.2-0.5=6.7hours[/tex] throughout your journey.

Hence, you can rest for  [tex]6.7-6=0.7hours=(0.7\times60)minutes=42minutes[/tex]

After driving 60 miles in the first hour and intending to arrive half an hour before your roommate who travels at 50 mph, you can afford a break time of 30 minutes at the rest stop.

The subject of your question is

relative speed

and

time management

, which falls under Mathematics. Your overall travelling speed is 60 mph. After driving for an hour, you decide to take a break at a rest stop. In that one hour, you've driven 60 miles, so there are 300 miles left to the destination.  Your roommate, who doesn't stop for a break, continues to drive at 50 mph. You want to arrive half an hour before she does, which means you essentially have her travel time less 30 minutes to reach the destination. Because her speed in relation to the remaining distance is constant, her remaining travel time is 300 miles divided by 50 mph which equals to 6 hours. So, you actually have 6 - 0.5 = 5.5 hours to travel the remaining 300 miles including your rest stop. Considering your speed of 60 mph, it will take you 300/60 = 5 hours to reach the destination. Therefore, the length of the break you can take while still beating your roommate to the destination is 5.5 hours minus 5 hours, which is

30 minutes

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Answer:

12.56 A.

Explanation:

The magnetic field of a conductor carrying current is give as

H = I/2πr ............................... Equation 1

Where H = Magnetic Field, I = current, r = distance, and π = pie

Making I the subject of the equation,

I = 2πrH............... Equation 2

Given: H = 1 T, r = 2 m.

Constant: π = 3.14

Substitute into equation 2

I = 2×3.14×2×1

I = 12.56 A.

Hence, the magnetic field = 12.56 A.

Answer:

0.019 atm

Explanation:

Assume ideal gas, so PV/T is constant where P is pressure, V is volume and a product of radius R cubed and a constant C, T is the temperature

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]P_2 = P_1\frac{V_1}{V_2}\frac{T_2}{T_1}[/tex]

[tex]P_2 = P_1\frac{CR_1^3}{CR_2^3}\frac{T_2}{T_1}[/tex]

[tex]P_2 = P_1\left(\frac{R_1^3}{R_2^3}\right)^3\frac{T_2}{T_1}[/tex]

[tex]P_2 = 1\left(\frac{0.9}{3}\right)^3\frac{210}{298}[/tex]

[tex]P_2 = 0.3^3*0.7 = 0.019 atm[/tex]

The student's question about the change in pressure of a helium-filled weather balloon as it expands and cools at altitude can be answered using the ideal gas law. By comparing initial and final conditions of pressure, volume, and temperature, and using the equation P2 = P1V1T2 / (T1V2), the new pressure can be determined.

The student is asking how to determine the pressure inside a helium-filled weather balloon when it rises to an altitude where the external conditions change. We are given the initial temperature, pressure, and radius of the balloon at liftoff, and the radius at its airborne location where the external temperature has decreased. To solve this problem, the ideal gas law is used in combination with the assumption that the balloon expands isotropically (uniformly in all directions). Since the internal pressure of the balloon must balance the external air pressure plus the pressure due to the tension in the balloon's material, we need to use a modified version of the ideal gas law that accounts for changes in temperature and volume to find the new pressure.

Given that the temperature and volume of the balloon change upon reaching altitude, if the volume and temperature of a gas are changed and the amount of gas (number of moles) remains constant, the ideal gas law (PV = nRT) can be rearranged to show the relationship between initial and final states:

P1V1/T1 = P2V2/T2

Where P1, V1, and T1 are the initial pressure, volume, and temperature and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. Since we know all variables except for P2, we can solve for P2 by rearranging the equation to:

P2 = P1V1T2 / (T1V2)

However, it is important to note that we must convert the volumes from radius measurements to actual volumes using the formula for the volume of a sphere, V = (4/3)πr3, and we must use absolute temperatures in Kelvin.

Using this equation with the provided values (making sure to convert units where necessary), the student will be able to determine the pressure at the airborne location for the weather balloon.

Answer:

T = 98.1 N

Explanation:

Given:

- mass of bar m = 10 kg

- Length of the bar L = 2 m

- Angle between Cable and wall Q = 30 degres

Find:

- Find the tension in the cable.

Solution:

- Take moments about intersection of bar and wall, to be zero (static equilibrium)

                                    (M)_a = 0

                                    T*sin( 30 )*L - m*g*L/2 = 0

                                    T*sin(30) - m*g / 2 = 0

                                    T = m*g / 2*sin(30)

                                    T = 10*9.81 / 2*sin(30)

                                    T = 98.1 N

Answer:

0.00000131569788654 C

0

Explanation:

R = Radius = 7.1 cm

L = Length of shell = 251 cm

r = 25.2 cm

E = Electric field = 37400 N/C

Electric field is given by

[tex]E=\dfrac{2kq}{rL}\\\Rightarrow q=\dfrac{ErL}{2k}\\\Rightarrow q=\dfrac{37400\times 0.252\times 2.51}{2\times 8.99\times 10^{9}}\\\Rightarrow q=0.00000131569788654\ C[/tex]

The net charge on the shell is 0.00000131569788654 C

Here, 4.07<7.1 cm which means r'<R

From Gauss law the electric at that point is 0

Answer:

The force on the electron will become zero.

Explanation:

As electron has negative charge and proton has positive charge. The magnitude of charge on both particles is equal. Therefore, there will be a force of attraction between electron and proton. When another electron is brought near the proton the net charge in that area will become equal to zero. Therefore, first electron will not experience any force.

Final answer:

The original force between the first electron and proton remains unchanged when a second electron is placed next to the proton, but the first electron will experience an additional repulsive force from the second electron. This alters the net force acting on the first electron but does not directly change the force between it and the proton.

Explanation:

The question inquires about the effect on the force on an electron when a second electron is placed next to a proton, with all particles separated by a distance of 1 meter.

According to Coulomb's Law, which governs the electrostatic force between two charged particles, the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

When a second electron is placed next to the proton, this setup introduces additional forces: a repulsive force between the two electrons and an attractive force between the new electron and the proton.

However, the original question seems to focus on how the force on the first electron changes with this new arrangement. The presence of a second electron does not alter the magnitude of the force between the first electron and the proton directly since the forces here are evaluated pairwise, and the distance between them remains unchanged.

What changes, though, is the overall electrical environment, introducing a new set of forces that need to be considered for the complete system. The addition of the second electron introduces a repulsive force on the first electron, which is separate from but concurrent with the attractive force from the proton.

It is important to consider the net force acting on any charge in such scenarios, which would involve adding vectorially the attractive force from the proton and the repulsive force from the second electron.

Thus, while the force due to the proton-electron pair remains constant, the first electron experiences an additional repulsive force due to the second electron, affecting the net force on it but not the force between it and the proton directly.

Answer:

(a) 11.66 square meters per liter

(b) 11657.8 per meters

(c) 0.00211 gal per square feet

Explanation:

(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L

(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m

(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3

Answer:

a)The keys were thrown with an initial velocity of 10.0 m/s.

b)The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

Explanation:

Hi there!

The equations for the height and velocity of the keys are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h =height of the keys after a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive)

v = velocity of the keys at time t.

a) We know that at t = 1.50 s, h = 4.00 m and let´s consider that the origin of the frame of reference is located at the point where the keys are thrown so that h0 = 0. Then, using the equation of height, we can obtain the initial velocity.

h = h0 + v0 · t + 1/2 · g · t²

4.00 m = v0 · 1.50 s - 1/2 · 9.81 m/s² · (1.50 s)²

4.00 m + 1/2 · 9.81 m/s² · (1.50 s)² = v0 · 1.50 s

v0 = ( 4.00 m + 1/2 · 9.81 m/s² · (1.50 s)²) / 1.50 s

v0 = 10.0 m/s

The keys were thrown with an initial velocity of 10.0 m/s.

b) Now, using the equation of velocity we can calculate the velocity at t = 1.50 s:

v = v0 + g · t

v = 10.0 m/s - 9.81 m/s² · 1.50 s

v = -4.72 m/s

The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

Answer:

[tex]t=\dfrac{d}{v_0cos(\theta )}[/tex]

Explanation:

The background information:

A student throws a water balloon with speed v0 from a height h = 1.8m at an angle θ = 29° above the horizontal toward a target on the ground. The target is located a horizontal distance d = 9.5 m from the student’s feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position.

The time it takes for the balloon to reach the target is equal to the target distance [tex]d[/tex] divided by the horizontal component of the velocity [tex]v_0[/tex]:

[tex]t=\dfrac{d}{v_x}[/tex]

where [tex]v_x[/tex] is the horizontal component of the velocity [tex]v_0[/tex], and it is given by

[tex]v_x=v_0cos(\theta)[/tex];

Therefore, we have

[tex]\boxed{t=\dfrac{d}{v_0cos(\theta)} }[/tex]

The time it takes for the balloon to reach the target can be found by solving the equation x = xo + vot + at². By rearranging the equation and substituting the given values, the time can be determined. For example, if the balloon has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude, it will spend 3.79 s in the air.

The time it takes for the balloon to reach the target, denoted by t, can be determined by the vertical motion of the balloon. The equation x = xo + vot + at² can be used to solve for t, since the only unknown in the equation is t.

By rearranging the equation and substituting the given values, we can solve for t.

For example, if the balloon has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude, it will spend 3.79 s in the air.

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Answer: the force exerted by the book on the earth

The reaction force to the force of gravity acting on a book resting on a shelf is the force exerted by the shelf, also known as the normal force. This force counteracts the force of gravity and is equal to the weight of the book. Understanding this concept requires knowledge of Newton's Second and Third Laws of Motion.

The reaction force to the force of gravity acting on a book resting on a shelf is the Force exerted by the shelf on the book. The shelf applies a Normal force, perpendicular to its surface, that counteracts the force of gravity pulling down on the book. This is directly tied into Newton's Third Law, which states that for every action, there is an equal and opposite reaction.

The normal force exerted by the shelf is a type of contact force and is exactly equal to the weight of the book (which is the force of gravity acting on the book). If the normal force were weaker, the book would begin to sink into the shelf. If it were stronger, the book would start to lift off of the shelf. Frictional force also plays a role here, preventing the book from sliding off the shelf, but it is not the reaction force to gravity in this case. Newton's Laws of Motion, particularly the Second and Third Laws, are key to understanding these dynamics.

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Answer:

The maximum volume flow rate is 0.03745m^3/s

Explanation:

Power input (Pi) = 9-hp = 9×746W = 6,714W

Elevation (h) = 15m

Efficiency (E) = 82% = 0.82

Density (D) of water = 1000kg/m^3

E = Po/Pi

Po (power output) = E×Pi = 0.82×6,714W = 5505.48W

Po = mgh/t

m/t (mass flow rate) = Po/gh = 5505.48/9.8×15 = 37.45kg/s

Volume flow rate = mass flow rate ÷ density = 37.45kg/s ÷ 1000kg/m^3 = 0.03745m^3/s

To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically

[tex]KE = PE[/tex]

[tex]\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}[/tex]

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that

[tex]q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}[/tex]

Replacing,

[tex]q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}[/tex]

[tex]q_2 = 3.6531nC[/tex]

Therefore the charge on the sphere is 3.6531nC

To solve this problem we will apply the concepts of the potential difference such as the product between the electric field and the distance, then we will use two definitions of capacitance, the first depending on the Area and the second depending on the load to find the Area. Finally we will look for capacitance with the values already obtained in the first sections of this problem

PART A) Potential Difference is

[tex]V = Ed[/tex]

Here,

E = Electric Field

d = Distance

Replacing,

[tex]V = (5*10^6)(3.0*10^{-3})[/tex]

[tex]V = 15000V= 15kV[/tex]

PART B) Capacitance of the capacitor is

[tex]C = \frac{\epsilon_0 A}{d}[/tex]

Here,

A = Area

[tex]\epsilon_0[/tex] = Permittivity Vacuum

d = Distance

Rearranging to find the Area we have,

[tex]A = \frac{Cd}{\epsilon_0}[/tex]

We know at the same time that Capacitance is the charge per Voltage, then

[tex]C = \frac{Q}{V}[/tex]

Replacing at this equation we have that

[tex]A = \frac{Qd}{V\epsilon_0}[/tex]

[tex]A = \frac{(79*10^{-9})(3*10^{-3})}{(15000)(8.853*10^{-12})}[/tex]

[tex]A = 1.78mm^2[/tex]

PART C)

Capacitance is given by,

[tex]C = \frac{Q}{V}[/tex]

[tex]C =\frac{79*10^{-9}}{15000}[/tex]

[tex]C =5.26pF[/tex]

The batteries do not make the charges move as much as the hand generator. This is because batteries store energy chemically, and depending on their properties, they may not provide enough energy to power all the bulbs in the circuit. In contrast, hand generators can generate a higher voltage due to a greater induced EMF from rapid movement in a magnetic field.

The best explanation for this phenomenon is that 'the batteries do not make the charges move as much' as compared to the hand generator. Hand generators work on the principle of electromagnetic induction where the rapid movement of a coil in a magnetic field can induce a substantial electromotive force (emf). This high emf allows the generator to power more light bulbs in a circuit. This is because the faster the generator is spun, the greater the induced emf, which in turn produces a higher voltage.

On the other hand, the batteries store energy as a chemical reaction waiting to happen, not as electric potential. This reaction only runs when a load is attached to both terminals of the battery. Hence, depending on the energy storage and discharge properties of the battery used, they may not be able to provide enough energy to sufficiently power all the bulbs in a circuit, especially if they are made to power more than one bulb simultaneously. Hence, in the case of the two batteries, they are not able to cause the charges to move as much as the hand generator, and this results in a lower voltage and consequently dimmer lights.

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Answer:

Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars.

False

Explanation:

The “hot Jupiters” joint word startes to be used to be able to describe planets like 51 Pegasi b, a planet with a 10-day-or-less orbit and a mass 25% or greater than Jupitere, circling a sun-like star planet in 1995, which was found by astronomers Michel Mayor and Didier Queloz, who were awarded the 2019 Nobel Prize for Physics along with the cosmologist James Peebles for their “contributions to our understanding of the evolution of the universe and Earth’s place in the cosmos.”

Now  we know a total of 4,000-plus exoplanets, but only a few more than 400 meet the definition of the enigmatic hot Jupiters which, tell us a lot about how planetary systems form, and what kinds of conditions cause extreme results.

In a 2018 paper in the Annual Review of Astronomy and Astrophysics, astronomers Rebekah Dawson of the Pennsylvania State University and John Asher Johnson of Harvard University reviewed on how hot Jupiters might have formed, and would be the meaning for the rest of the planets in the galaxy.

Answer:

[tex]Q=41.90kJ[/tex]

Explanation:

The heat lost by the water in the cooling process is transferred to the muscles. Therefore, we must calculate this water lost heat, which is defined as:

[tex]Q=mc\Delta T[/tex]

Where m is the water's mass, c is the specific heat capacity of the water and [tex]\Delta T=T_f-T_0[/tex] is the change in temperature. Replacing the given values:

[tex]Q=715g(4186\frac{J}{g^\circC}})(51^\circ C-37^\circ C)\\Q=41901.86J\\Q=41.90kJ[/tex]

To solve this problem we will rely on the theorems announced by Newton and Coulomb about the Gravitational Force and the Electrostatic Force respectively.

In the case of the Force of gravity we have to,

[tex]F_g = G\frac{m_pm_e}{d^2}[/tex]

Here,

G = Gravitational Universal Constant

[tex]m_p[/tex] = Mass of Proton

[tex]m_e[/tex] = Mass of Electron

d  = Distance between them.

[tex]F_g = (6.673*10^{-11} kg^{-1} \cdot m^3 \cdot s^{-2}) (\frac{(1.672*10^{-27}kg)(9.109*10^{-31})}{(52.9pm)^2})[/tex]

[tex]F_g = 3.631*10^{-47}N[/tex]

In the case of the Electric Force we have,

[tex]F_e = k\frac{q_pq_e}{d^2}[/tex]

k = Coulomb's constant

[tex]q_p[/tex] = Charge of proton

[tex]q_e[/tex] = Charge of electron

d = Distance between them

[tex]F_e = (9*10^9N\cdot m^2 \cdot C^{-2})(\frac{(1.602*10^{-19}C)(1.602*10^{-19}C)}{(52.9pm)^2})[/tex]

[tex]F_e = 82.446*10^{-9}N[/tex]

Therefore

[tex]\frac{F_e}{F_g} = 2.270*10^{39}[/tex]

We can here prove that the statement is True

It has been proved in below calculation that the gravitational force exerted by a proton on an electron is [tex]2\times10^{39}[/tex]times weaker than the electric force that the proton exerts on an electron

Gravitational force is the universal force of attraction acting between two bodies.

It is given by,

[tex]Fg=G\times\dfrac{m_1 \times m_2}{x^2}[/tex]

Here [tex]G[/tex] is gravitational constant [tex]m[/tex] is the mass of the bodies and [tex]x[/tex] is the distance between bodies.

Electric force is the force of attraction or repulsion acting between two charged bodies,

[tex]F_E=k\times\dfrac{q_1 \times q_2}{x^2}[/tex]

Here, [tex]k[/tex] is Coulomb's constant [tex]q[/tex] is the charge and [tex]x[/tex] is the distance between bodies.

The gravitational force exerted by a proton on an electron is,

[tex]Fg=6.673\times{10^{-11}}\times\dfrac{1.673\times10^{-27}\times9.1094\times10^{-31}}{x^2}[/tex]

The electric force exerted by a proton on an electron is,

[tex]F_E=9\times{10^{9}}\times\dfrac{1.602\times10^{-19}\times1.602\times10^{-19}}{x^2}[/tex]

Compare both,

[tex]\dfrac{F_g}{F_e} =\dfrac{6.673\times{10^{-11}}\times\dfrac{1.673\times10^{-27}\times9.1094\times10^{-31}}{x^2}}{9\times{10^{9}}\times\dfrac{1.602\times10^{-19}\times1.602\times10^{-19}}{x^2}}\\\dfrac{F_g}{F_e} =\dfrac{1}{2\times10^{39}}[/tex]

Thus, It has been proved in below calculation that the gravitational force exerted by a proton on an electron is [tex]2\times10^{39}[/tex]times weaker than the electric force that the proton exerts on an electron.

Learn more about the gravitational force electric force here;

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Answer:

9.09*10^-13

Explanation:

Certain values of the problem where omitted,however the omitted values were captured in the solution.

Step1:

Avogadro's number (NA) I = 6.02*10^23 atoms/mole.

Step2:

To determine the number of moles of copper that are present, thus: Using the mass and atomic mass :

n = m/A

n = 50.0g/63.5g/mol

Therefore, since the are 29 protons per atom, I the number of protons can be determined as follows :

Np = nNA*29 protons /atom

Np=(50.0gm/63.5g/mol)(6.02*10^23 atoms/mol) * (29 protons / C u atom)

Np= 1.375*10^25 protons

Note that there are same number of electrons as protons in a neutral atom, I therefore the removal of electrons to give the copper a net change, hence the result is 1.375*10^25

Step3:

To determine the number electrons , removed to leave a net charge of 2.00Uc, then remove -2 .00Uc of charge, so that the number of electrons to be removed are as follows :

Ne(removed)=

Q/qe= -2.00*10^-6c/-1.60* 10^-19c

Ne(removed)=1.25*10^13 electrons removed

Step4:

To calculate the fraction of copper's electron by taking the ratio of the number of electrons initially present:

Ne,removed/Ne,initially=1.25*10^13/ 1.37*10^25 = 9.09*10^-13

Answer:

F= 134.92 N

Explanation:

Given that

The mass of the moon ,M = 7.4 x 10²² kg

The mass of the man ,m = 79 kg

The radius ,R= 1.7 x 10⁶ m

The force exerted by moon is given as

[tex]F=G\dfrac{Mm}{R^2}[/tex]

Now by putting the values in the above equation we get

[tex]F=6.67\times 10^{-11}\times \dfrac{79\times 7.4\times 10^{22}}{(1.7\times 10^6)^2}\ N\\F=134.92 N[/tex]

Therefore the force will be 134.92 N.

F= 134.92 N

Answer:

[tex]V=\sqrt{V_{0}^{2}+2gy}[/tex]

Explanation:

Data given,

[tex]velocity,v =v_{0}\\ angle =\alpha _^{0}[/tex]

since the motion part is describe by a projectile motion, the acceleration along the horizontal axis is zero

Hence using the equation v=u+at we have the following equation ,

the velocity along the horizontal axis to be

[tex]V_{x}=V_{0}cos\alpha _{0} \\[/tex]

the velocity along the vertical axis to be

[tex]V_{y}=V_{0}sin\alpha _{0}-gt \\[/tex]

the magnitude of this velocity can be determine using Pythagoras theorem

[tex]V^{2}=V_{x}^{2} +V_{y} ^{2}[/tex]

if we substitute the expressions we have  

[tex]V^{2}=V_{0}^{2}cos\alpha _{0}^{2} +(V_{0}sin\alpha _{0}-gt)\\expanding \\V^{2}=V_{0}^{2}(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})-2gtsin\alpha _{0}+(gt)^{2}\\(cos\alpha _{0}^{2}+sin\alpha _{0}^{2})=1\\V^{2}=V_{0}^{2}-2gtsin\alpha _{0}+(gt)^{2}\\[/tex]

[tex]V^{2}=V_{0}^{2}-2gtV_{0}sin\alpha _{0}+(gt)^{2}\\V^{2}=V_{0}^{2}-2g(V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2} )\\V_{0}sin\alpha _{0}-\frac{1}{2}gt^{2}=distance=y\\V^{2}=V_{0}^{2}-2gy\\ for upward \\V^{2}=V_{0}^{2}+2gy\\V=\sqrt{V_{0}^{2}+2gy}[/tex]

The speed of the rock, just before it strikes the ground (maximum speed) can be given as,

[tex]v=\sqrt{v_0^2-2gh} \\[/tex]

Witch is independent to angle [tex]a_0[/tex].

The speed of the object falling from a height achieved the maximum speed, just before hitting the ground.

Given information-

The rock is thrown with a velocity [tex]v_0[/tex].

The rock is thrown with a angle of [tex]a_0[/tex].

The height of the building is [tex]h[/tex].

The height of the building can be given as,

[tex]h=v_0\times\sin(a_0)-\dfrac{1}{2}gt[/tex] .........1

Let the above equation as equation 1,

The horizontal velocity of the rock can be given as,

[tex]v_h=v_0\times\cos (a_0)[/tex].

The vertical velocity of the rock can be given as,

[tex]v_v=v_0\times\sin(a_0)-gt[/tex]

Here, [tex]g[/tex] is the gravitational force and [tex]t[/tex] is time.

Now the magnitude of the velocity can be given as,

[tex]v=\sqrt{v_h^2+v_v^2} \\v=\sqrt{(v_0\times\cos(a_0))^2+(v_0\times\sin(a_0)-gt)^2} \\v=\sqrt{v_0^2\times\cos^2(a_0)+v_0^2\times\sin(a_0)^2+(gt)^2-2gtv_0\times\sin(a_0)} \\v=\sqrt{v_0^2(\cos^2(a_0)+\times\sin(a_0)^2)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\v=\sqrt{v_0^2(1)-2g(v_0\times\sin(a_0)-\dfrac{1}{2}gt^2} \\[/tex]

Put the values of [tex]h[/tex] from equation 1 to the above equation as,

[tex]v=\sqrt{v_0^2-2gh} \\[/tex]

Hence the speed of the rock, just before it strikes the ground (maximum speed) can be given as,

[tex]v=\sqrt{v_0^2-2gh} \\[/tex]

Witch is independent to angle [tex]a_0[/tex].

Learn more about the velocity of free falling body here;

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Using physics principles and the kinematic equation, the problem calculates the stopping distance of a car decelerating from 80 mi/hr, based on known stopping distance at 60 mi/hr. It involves converting speeds and applying algebraic manipulation to solve for the new distance.

To solve the problem of finding the stopping distance from an initial speed of 80 mi/hr, given the stopping distance from 60 mi/hr is 120 ft, we use the principle of physics that relates deceleration, distance, and speed. This approach requires converting speeds from miles per hour to feet per second, applying the kinematic equation v² = u² + 2as, and solving for the unknown stopping distance.

First, speeds are converted from miles per hour to feet per second. Given the initial situation: a car decelerates from 60 mi/hr to rest over 120 feet. Converting 60 mi/hr to feet per second gives 88 feet per second (using 1 mi = 5280 feet and 1 hour = 3600 seconds). Applying v² = u² + 2as (where v is final speed, u is initial speed, a is acceleration, and s is stopping distance) allows us to calculate the deceleration using the initial conditions. Next, using the same deceleration, we calculate the stopping distance from 80 mi/hr (converted to feet per second).

The detailed calculation involves algebraic manipulation of the kinematic equation to solve for the new stopping distance using the derived constant deceleration from the 60 mi/hr case. It demonstrates how a car's stopping distance increases with a square of the speed, illustrating the critical relationship between speed, stopping distance, and safety.

The stopping distance from an initial speed of 80 mi/hr would be approximately 195.555 ft, assuming the same constant deceleration as the first scenario.

To find the stopping distance from an initial speed of 80 mi/hr, we first need to determine the acceleration of the car during braking. Using the initial velocity of [tex]60 mi/hr[/tex] and the stopping distance of 120 ft, we can calculate the acceleration using the kinematic equation:

[tex]\[ v_f^2 = v_i^2 + 2as \][/tex]

Given that [tex]\( v_f = 0 \)[/tex] (the car comes to rest), [tex]\( v_i = 60 \) mi/hr, and \( s = 120 \)[/tex]ft, we rearrange the equation to solve for [tex]\( a \)[/tex]:

[tex]\[ a = \frac{{v_f^2 - v_i^2}}{{2s}} \][/tex]

Substituting the values:

[tex]\[ a = \frac{{0 - (60 \, \text{mi/hr})^2}}{{2 \times 120 \, \text{ft}}} \]\[ a \approx -33.333 \, \text{ft/s}^2 \][/tex]

Now, using this acceleration value, we can find the stopping distance[tex](\( s \))[/tex] from an initial speed of 80 mi/hr. With [tex]\( v_i = 80 \) mi/hr[/tex], we convert it to feet per second and use it in the same kinematic equation:

[tex]\[ v_i = 80 \, \text{mi/hr} \times 1.46667 \, \text{ft/s/mi/hr} = 117.333 \, \text{ft/s} \]\[ s = \frac{{-(117.333 \, \text{ft/s})^2}}{{2 \times (-33.333 \, \text{ft/s}^2)}} \]\[ s \approx 195.555 \, \text{ft} \][/tex]

Therefore, the stopping distance from an initial speed of [tex]80 mi/hr[/tex]would be approximately 195.555 ft, assuming the same constant deceleration as the first scenario.