If the water in a lake is everywhere at rest, what is the pressure as a function of distance from the surface? The air above the surface of the water is at standard sea-level atmospheric conditions. How far down must one go before the pressure is 1 atm greater than the pressure at the surface?

Final answer:

The acceleration due to gravity at the altitude of the Columbia shuttle is approximately 8.95 m/s^2, which is about 91% of Earth's surface gravity.

Explanation:

To determine the acceleration due to gravity at the altitude at which Columbia is orbiting, we apply the universal law of gravitation, which is represented by the formula:

F = G Mm/r²

From this formula, we can derive the expression for acceleration due to gravity (g):

g = GM/r²

Given:

Radius of Earth (rE) = 6.4 × 106 m

Altitude of Columbia (h) = 0.3 × 106 m

Mass of Earth (M) = 6.0 × 1024 kg

Gravitational constant (G) = 6.67408 × 10-11 m3kg-¹s-²

Total distance from the center of Earth (r) = rE + h

The resulting equation for the acceleration due to gravity at the altitude of Columbia is:

g = (6.67408 × 10-11 m3kg-1s-2) × (6.0 × 1024 kg) / (6.7 × 106 m)²

Upon calculating, this yields an acceleration due to gravity of approximately 8.95 m/s². This acceleration is about 91% of the gravitational acceleration at Earth's surface, which is 9.8 m/s².

Final answer:

The percent uncertainty of a blood pressure measurement of 120 mm Hg with an uncertainty of ±2 mm Hg is 1.67%. Using this percentage, the uncertainty in a blood pressure measurement of 71 mm Hg would be ±1.2 mm Hg, after rounding to one decimal place.

Explanation:

To calculate the uncertainty in a blood pressure measurement of 71 mm Hg, given the percent uncertainty is the same as that for a measurement of 120 mm Hg with an uncertainty of ± 2 mm Hg, we first find the percent uncertainty of the original measurement.

The percent uncertainty is calculated by dividing the uncertainty by the actual measurement and then multiplying by 100 to get a percentage:

Percent Uncertainty = (Uncertainty / Measurement) × 100

For the measurement of 120 mm Hg, the percent uncertainty is:

(2 mm Hg / 120 mm Hg) × 100 = 1.67%

Assuming the same percent uncertainty for a blood pressure measurement of 71 mm Hg, you would use the same percentage to find the uncertainty:

Uncertainty = Percent Uncertainty × Measurement / 100

Uncertainty in 71 mm Hg = 1.67% × 71 mm Hg / 100 = 1.187 mm Hg

Since we typically round to the nearest whole unit or significant digit the uncertainty would be ± 1.2 mm Hg (rounded to one decimal place).

To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.

According to the graph, the sum of the horizontal components would be

[tex]\sum T_x = 0[/tex]

[tex]TSin35\° -TSin35\° = 0[/tex]

[tex]0 = 0[/tex]

The sum of the vertical components is

[tex]\sum T_y = 0[/tex]

[tex]2TCos35\° - mg = 0[/tex]

[tex]2TCos35\° = 600N[/tex]

[tex]T = \frac{600N}{2Cos35\°}[/tex]

[tex]T = 366.2N[/tex]

Therefore the net force on the new vine is 366.2N

Answer:

as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Explanation:

When you rub a foam rod that is an insulator with felt it acquires some loads, when separated these two the barrel is loaded

When approaching the loaded rubber rod, suppose positively, to the aluminum ball that is a metal, the positive charge attracts the free electors of the aluminum until a balance of the electric force and the tension of the wire in which it is hung is established The aluminum foil therefore the aluminum foil approaches the rubber ball, due to the electrostatic attraction, the closer the aluminum wave deflects.

When the ball touches the sheet, the free charges of the metal pass to the insulator neutralizing the existing charges, whereby the electric field and the electric force decrease. Consequently the aluminum sheet descending and moving away from the rubber ball.

 In summary, as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Due do the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

When the insulating material are rubbed against each other, the electric charge generate between them.

The effect of friction charge between rod and ball leads to,

Hence due to the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

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Explanation:

For the given question, distance is given on x-axis and force is given on y-axis. Both distance and force are showing an interval of 5 with starting point 0.

Hence, work in 5 m to 12.5 m is equal and opposite to 12.5 m to 20 m. Since, work is equal then it means that kinetic energy at 5 m will also be equal to the kinetic energy at 20 m.

Therefore, we can conclude that the kinetic energy of the 2 kg object when d equals 20 m is the same as when d is most nearly  5 m.

Final answer:

The velocity of the 2 kg object when d equals 20 m is most nearly 20.4 m/s.

Explanation:

The equation for kinetic energy is K = 1/2mv^2, where K is the kinetic energy, m is the mass, and v is the velocity of the object. To find the velocity when d equals 20m, we can use the equation K = 1/2mv^2 and rearrange it to solve for v. We know the mass of the object is 2kg and the kinetic energy is the same, so we can set up the equation as follows:

1/2(2kg)(v)^2 = 1/2(2kg)(20.4m/s)^2

Simplifying the equation, we find that v is approximately 20.4m/s. Therefore, when d equals 20m, the velocity of the object is most nearly 20.4m/s.

Answer:

Horizontal component of velocity will be 61.28 ft/sec

Vertical component of the velocity will be 51.423 ft/sec

Explanation:

We have given velocity of football v = 80 ft/sec'

Angle at which football is thrown [tex]\Theta =40^{\circ}[/tex]

Now we have to fond the horizontal and vertical component of the velocity

Horizontal component of velocity [tex]v_x=vcos\Theta =80\times cos40^{\circ}=80\times 0.766=61.28ft/sec[/tex]

Vertical component of the velocity [tex]v_y=vsin\Theta =80\times sin40^{\circ}=51.423ft/sec[/tex]

Answer:

Option-(D):Thermal energy unchanged, temperature decreased.

Explanation:

Thermal energy and Temperature,T:

Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.

While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = [tex]\frac{k*q}{r}[/tex]

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = [tex]\frac{k*q1}{x}[/tex] +[tex]\frac{k*q2}{(1.63m-x)}[/tex] = 0

⇒[tex]k*q1* (1.63m - x) = -k*q2*x[/tex]:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

Answer:

none

Explanation:

a. the unit of height does not carry a square( m^2 is unit of area)

b. time cannot be in m/s. it is unit of speed

c. m/s^2 are the units of acceleration

None of the options are correct on the basis of dimensional analysis.

(a)The height of the Transamerica Pyramid is 332 m².

the dimension of height is not correct, it must be m instead of m².

(b)The time duration of a fortnight is 66 m/s.

the dimension of time is s ( seconds) not m/s (meter/second).

(c) The speed of the train is 9.8 m/s².

The dimentsion of speed is m/s (meter/second) not m/s².

Hence, none of the options are correct.

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Answer:

New force between them will become [tex]\frac{1}{36}[/tex] times

Explanation:

Let the charge on both the object are [tex]q_1\ and\ q_2[/tex] and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]

According to question [tex]F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}[/tex]-----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force [tex]F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}[/tex]------------------eqn 2

Comparing eqn 1 and eqn 2

[tex]F_{new}=\frac{F}{36}[/tex]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

[tex]Q_H<0[/tex]

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

[tex]Q_C>0[/tex]

That is why the answer will be

[tex]Q_H<0[/tex] ,[tex]Q_C>0[/tex]

In the context of energy transfers with hot and cold reservoirs, the sign convention is

Given that there are two reservoir of energy. Sign convention for heat and work :

1) If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2) If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

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Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

[tex]\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s[/tex]

[tex]\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM[/tex]

Therefore the speed in RPM will be 62.64 RPM.

Answer:

Incomplete question.

Complete question given below

a) 2.7933 C/s

b) 714.45 s

Explanation:

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails {1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt.

(a) Calculate the rate of temperature increase in degrees Celsius per second  if the mass of the reactor core is  1.60105*10^6 kg and it has an average specific heat of  0.3349 kJ/kgº.

(b) How long would it take to obtain a temperature increase of  which could cause some metals holding the radioactive materials to melt?

Part a)

[tex]Q_{rate} = m*c*\frac{dT}{dt}\\\frac{dT}{dt} = \frac{Q_{rate}}{m*c} \\\\\frac{dT}{dt} = \frac{150*10^6}{1.6 * 10^5*334.9} \\\\\frac{dT}{dt} = 2.7993 C/s[/tex]

Part b)

[tex]dt = \frac{dT}{2.7993} \\\\dt = \frac{2000}{2.7993} \\\\dt = 714.45 s[/tex]

To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is

Given that,

[tex]M_E =5.97*10^24 kg[/tex]

[tex]m_m = 7.36*10^22 kg[/tex]

[tex]r_E = 6370 km[/tex]

[tex]r_m = 1737 km[/tex]

The distance between Earth and Moon

[tex]x = 384403*10^3 m[/tex]

As the net gravitational force is zero we have that

[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]

[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]

Replacing with our values

[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]

[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]

[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]

Replacing x as the distance betwen moon and earth,

[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]

Solving the polynomial we have that

[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]

So the minimum value is [tex]3.4506*10^{8}m[/tex]

The force constant of the spring is 1.932 N/m. The maximum mass of the glider can be calculated using Newton's second law. The magnitude of the maximum acceleration of the glider can be calculated using the equation for acceleration in SHM.

In simple harmonic motion (SHM), the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Let's calculate the force constant of the spring by using Hooke's Law. We know that the force exerted by the spring is given by F = kx, where F is the force, k is the force constant, and x is the displacement. At the maximum displacement, the force exerted is F = k * 2.05 m. At the minimum displacement, the force exerted is F = k * 1.06 m. Equating these two forces, we get k * 2.05 = k * 1.06. Solving this equation, we find the force constant of the spring to be k = 2.05 / 1.06 = 1.932 N/m.

The maximum mass of the glider can be found by calculating the maximum force that can be exerted by the spring. The maximum force is given by F = k * x, where F is the force, k is the force constant, and x is the maximum displacement. Plugging in the values, we get F = 1.932 N/m * 2.05 m = 3.961 N. Using Newton's second law, F = m * a, where F is the force, m is the mass, and a is the acceleration, we can calculate the maximum mass. Rearranging the equation gives m = F / a. Since the acceleration in SHM is given by a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period, we can plug in the given values to find the maximum mass.

The magnitude of the maximum acceleration of the glider can be found by using the equation a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period. Plugging in the given values, we can calculate the magnitude of the maximum acceleration.

Answer:

25.1%

Explanation:

Let's say m is the true weight and M is the weight of the counterweight.

The original distance between the pans and the center is 25.1 cm / 2 = 12.55 cm.  The fulcrum is then moved 1.4 cm towards the counterweight.

Sum of the moments about the fulcrum:

m (12.55 + 1.4) − M (12.55 − 1.4) = 0

13.95 m − 11.15 M = 0

13.95 m = 11.15 M

M / m = 13.95 / 11.15

M / m = 1.251

The true weight is being marked up by 25.1%.

The true weight of the goods being marked up by the shopkeeper is 25.1%.

A lever is a simple machine made up of a beam or rigid rod that pivots at fulcrum. A lever is a rigid body that can rotate on a single point. The lever is classified into three types based on the locations of the fulcrum, load, and effort.

Let the true mass is m.

and  the reading f the mass is M.

The initial distance between the two pan is 25.1 cm.

So, The initial distance between the pan and the fulcrum is  = 25.1 cm / 2 = 12.55 cm.

The fulcrum is moved 1.4 cm towards the counterweight by the dishonest  shopkeeper.

At equilibrium: the moment of force of two hands from fulcrum is equal. Hence,

mg (12.55 + 1.4)  = Mg (12.55 − 1.4)

13.95 m = 11.15 M

M / m = 13.95 / 11.15

M / m = 1.251

Hence, the true weight is being marked up by = {(M-m)/m}×100%

= {(1.251-1)×100%}

= 25.1%.

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Explanation:

Given that,

Mass of the raindrops, [tex]m=0.5\ g=0.0005\ kg[/tex]

Charges, [tex]q=1\ mC=10^{-3}\ C[/tex]

Distance between raindrops, d = 1 cm = 0.01 m

(a) The force due to motion of raindrop is balanced by the electric force between charges. It is given by :

[tex]ma=\dfrac{kq^2}{r^2}[/tex]

a is the acceleration of the raindrop

[tex]a=\dfrac{kq^2}{r^2m}[/tex]

[tex]a=\dfrac{9\times 10^9\times (10^{-3})^2}{(0.01)^2\times 0.0005}[/tex]

[tex]a=1.8\times 10^{11}\ m/s^2[/tex]

(b) It is clear that the acceleration is too large. It can break apart the rain drop.

(c) If the charge is of the factor of [tex]10^{-8}\ C[/tex], the charge would be more reasonable.            

Hence, this is the required solution.  

To find the acceleration of the charged raindrops, we can use Coulomb's Law and Newton's second law. The acceleration is found to be 1.8 x 10^13 m/s^2, which seems unreasonable for raindrops. The assumption of the raindrops acquiring charges of 1.00 mC is likely responsible for this result.

To calculate the acceleration of the two charged raindrops, we can use Coulomb's Law, which states that the force between two charges is given by: F = (k*q1*q2)/r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. The force experienced by each raindrop is equal in magnitude but opposite in direction.

The acceleration of each raindrop can be found using Newton's second law: F = m*a, where F is the net force, m is the mass of the raindrop, and a is the acceleration. Rearranging the equation, we have: a = F/m. Substituting the values into the equations and solving for acceleration, we find: a = (k*q^2)/(m*r^2). Plugging in the values, we have: a = (9 x 10^9 N*m^2/C^2 * (1 x 10^-3 C)^2) / (0.500 g * 10^-3 kg/g * (1 x 10^-2 m)^2). After calculating the values, we find that the acceleration of the two raindrops due to their charges is 1.8 x 10^13 m/s^2.

(b) What is unreasonable about this result? The result seems unreasonable because the acceleration is extremely high. Normally, objects with masses like raindrops would experience much lower accelerations. (c) The premise or assumption responsible for this result is that the raindrops have acquired charges of 1.00 mC, which is an extremely high charge for raindrops and likely not possible in reality.

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Answer:

The time it would take to sail around the world in a sailboat is 1000 hour.        

Explanation:

It is given that,    

The circumference of the Earth, [tex]d=10^4\ mile[/tex]

The speed of a sailboat, [tex]v=10\ mile/hour[/tex]

We need to find the time it would take to sail around the world in a sailboat. It can be calculated using the definition of velocity. It is given by :

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

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

[tex]t=\dfrac{10^4\ mile}{10\ mile/hour}[/tex]

t = 1000 hour

So, the time it would take to sail around the world in a sailboat is 1000 hour. Hence, this is the required solution.

Explanation:

Given:

source temperature T1= 825° C = 1099 K

Sink temperature T2= 15°C = 288 K

Heat Supplied to the engine Qs= 3200 KW

a) Efficiency of a Carnot engine is

[tex]\eta =1-\frac{T_1}{T_2}[/tex]

[tex]=1-\frac{1099}{288}[/tex]

=0.7379

=73.79%

b)

[tex]\eta= \frac{Power delivered}{power recieved}[/tex]

let W be the powe delivered

[tex]0.7379= \frac{W}{3200}[/tex]

W= 2361.408 KW

c) Heat rejected to the cold temperature Qr= Qs-W

= 3200-2361.408=838.60 KW

d) Entropy change in the sink

[tex]\Delta S= \frac{Qr}{T_{sink}}[/tex]

=838.60/288= 2.911 W/K

Answer:

a. [tex]\eta_{th}=0.7377=73.77\%[/tex]

b. [tex]P=2360655.7377\ W[/tex]

c. [tex]Q_o=839344.2622\ W[/tex]

d. [tex]\Delta s=2914.3897\ J.K^{-1}[/tex]

Explanation:

Given:

a.

Since the engine is a Carnot engine:

[tex]\eta_{th}=1-\frac{T_L}{T_H}[/tex]

[tex]\eta_{th}=1-\frac{288}{1098}[/tex]

[tex]\eta_{th}=0.7377=73.77\%[/tex]

b.

Now the power delivered:

since here we are given with the rate of heat transfer

[tex]P=\eta_{th}\times Q_i[/tex]

[tex]P=0.7377\times 3200000[/tex]

[tex]P=2360655.7377\ W[/tex]

c.

rate of heat rejection to the sink:

[tex]Q_o=Q_i-P[/tex]

[tex]Q_o=3200000-2360655.73770491808[/tex]

[tex]Q_o=839344.2622\ W[/tex]

d.

[tex]\Delta s=\frac{Q_o}{T_L}[/tex]

[tex]\Delta s=\frac{839344.2622}{288}[/tex]

[tex]\Delta s=2914.3897\ J.K^{-1}[/tex]

Answer:

A. [tex]\omega=11.1121\ rad.s^{-1}[/tex]

B. [tex]f=1.7685\ Hz[/tex]

C. [tex]T=0.5654\ s[/tex]

Explanation:

Given:

A)

for a spring-mass system the frequency is given as:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{56.8}{0.46}}[/tex]

[tex]\omega=11.1121\ rad.s^{-1}[/tex]

B)

frequency is given as:

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

[tex]f=\frac{11.1121}{2\pi}[/tex]

[tex]f=1.7685\ Hz[/tex]

C)

Time period of a simple harmonic motion is given as:

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

[tex]T=0.5654\ s[/tex]

Final answer:

Even if two forces are equal in magnitude and opposite in direction, they can still create a net torque if they are applied at different points that are at different distances from the object's pivot point.

Explanation:

In physics, particularly when discussing Newton's third law, it is essential to recognize that while two forces may be equal in magnitude and opposite in direction, they do not necessarily result in equilibrium if they are acting on different systems. When considering the possibility of a net torque, it is crucial to understand that torque is not just about the magnitude of the force, but also its point of application and the distance from the pivot point (or the axis of rotation).

To produce a torque, a force must be applied in such a manner that it causes the object to rotate around a pivot point. If two equal and opposite forces are applied at different points on an object, and those points are at different distances from the pivot point, then a net torque can occur.

For example, imagine a seesaw with equal forces pushing down on either end. If these forces are applied at different distances from the fulcrum (pivot point), the seesaw will rotate due to the net torque, despite the forces being equal and opposite. This demonstrates that even with equal and opposite forces, the effect on the object's rotational motion depends on the forces' points of application and their distances from the pivot point.

Answer:

Time period at the moon will be 2.828 sec

Explanation:

Let initially the length of the pendulum is l and acceleration due to gravity is g

And time period is given T = 2 sec

So time period of the pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]-------------eqn 1

Now on the planet acceleration due to gravity [tex]g_{new}=\frac{g}{2}[/tex]

So time period of the pendulum at this planet [tex]T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}[/tex]--------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}[/tex]

[tex]T_{new}=\sqrt{2}T=1.414\times 2=2.828sec[/tex]

So time period at the moon will be 2.828 sec

If the acceleration due to gravity decreases by a half, the period of a simple pendulum increases by √2, or approximately 1.41. Therefore, the period of the pendulum on the moon of Planet X, where gravity is half, would be 2.8 seconds.

The period of a simple pendulum in physics is influenced by its length and the acceleration due to gravity (g). It is calculated by the formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravity. To find the effect of a change in g on the period of the pendulum, we can look at the relationship between them in the formula. Because the formula for the period of a pendulum involves the square root of the gravity, if gravity is reduced by a half, the period will increase by a factor of √2, which is approximately 1.41.

Thus, if the pendulum's period on Planet X was 2.0 s, then on its moon, where gravity is halved, the period will be approximately 2.0 s * 1.41 = 2.8 s.

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Answer:

The speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle  (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}[/tex]

[tex]v=4.52\times 10^6\ m/s[/tex]

For proton,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}[/tex]

[tex]v=2468.02\ m/s[/tex]

So, the speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

Answer:

(a) The x component of the net electrostatic force on charge q3 is 0.195N

(b) The x component of the net electrostatic force on charge q3 is 0.021N

Explanation:

The force of q1 on q3 is a repulsive force and those of q2 and q4 on q3 are attractive forces. From second diagram at the bottom of the page in the attachment below it can be seen that the force of q1 on q3 is directed vertically downward and has on y component while the force of q4 on q3 is directed to the right and has on x component. The force of q2 on q3 has both x and y components. The full solution can be found in the attachment below.

Thank you for reading.

The electric force between two charged bodies at rest is called as electrostatic force.

The values are the following:

The x component of the net electrostatic force = 0.17 N

The y component of the net electrostatic force = 0.045 N

For component X

[tex]= (8.99 \times 10^{9} N.m2/ C^{2} \dfrac{(2.0\times10^{-7} )^2} {(0.050 m)^2}(-2 +\frac{2}{2\sqrt{2} } )\\ = 0.17 N[/tex]

X= 0.17 N

For component Y

Y = 0.045 N.

Hence, X= 0.17 N and Y = 0.045 N.

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Answer:

4009.21658986 Hz

Explanation:

v = Speed of sound in air = 348 m/s

L = Length of tube = 2.17 cm

Fundamental frequency for a tube where one end is closed and the other end open is given by

[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{348}{4\times 2.17\times 10^{-2}}\\\Rightarrow f=4009.21658986\ Hz[/tex]

The fundamental frequency of the canal is 4009.21658986 Hz

Answer: The answer is 258

Explanation: The distance between the nuclei of the atoms in a diatomic molecule X2 is measured to be 516pm

Therefore the atomic radius of X2 is

516/2=258

The main reasons for having few impact craters on Earth compared to moon is due to the presence of atmosphere, water bodies and tectonic activities.

Explanation:

Impact craters are formed on the surface of any object when another very high velocity object hits the surface of the particular object. As force exhibited by any object is directly proportional to the product of mass and acceleration of the object.

In case of moon or other natural satellites, there are no means of reducing the acceleration or mass of the incident high velocity object. But in earth, the presence of different layers of atmosphere causes sufficient friction to reduce the size as well as acceleration of the impeding high velocity object.

Then the presence of larger volume of water bodies on the Earth's surface reduces the chance of hitting on land areas. The third important factor reducing the impact craters are the variation in tectonic plates which helps in keeping only recent impacts on the Earth's surface and erase the old impacts.

So the presence of different layers of atmosphere, water bodies and tectonic plats are the main reasons for having fewer impact craters on Earth compared to the Moon.

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

[tex]\int\ {E} \, dA = E*A = E*\pi *R^{2}[/tex]

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

Final answer:

The electric flux through a hemisphere surface is calculated using Gauss's Law, where the electric field lines are parallel to the area element vector on the surface. The flux through a closed spherical surface is independent of its radius. Choosing a Gaussian surface with appropriate symmetry aids in evaluating the flux integral.

Explanation:

The magnitude of the electric flux through the hemisphere surface can be calculated using Gauss's Law. For a hemispherical surface of radius R in a uniform electric field,

Answer:

A: 5.67*10^-5 N

B: 3.49*10^-5 N

C: -9.16*10^-5 N

Explanation:

[tex]\\A\\F_{A} = F_{BA} + F_{CA} =G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(\frac{363*517}{0.5^2} + \frac{154*363}{0.75^2})\\\\= 5.67*10^-5 N \\\\B\\\\F_{B} = -F_{BA} + F_{CB} =-G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{b}}{r^2_{BC}} \\\\= (6.67*10^(-11))*(-\frac{517*363}{0.5^2} + \frac{517*154}{0.25^2})\\\\= 3.49*10^-5\\\\[/tex]

[tex]F_{C} = -F_{CA} - F_{CB} =-G*\frac{m_{b}*m_{c}}{r^2_{BC}} - G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(-\frac{517*154}{0.25^2} - \frac{154*363}{0.75^2})\\\\= -9.16*10^-5[/tex]

Answer:

binding energy will be [tex]0.1633\times 10^{-15}J[/tex]

Explanation:

We have given wavelength of photon [tex]\lambda =121pm=121\times 10^{-12}m[/tex]

Velocity of light [tex]c=3\times 10^8m/sec[/tex]

Plank's constant [tex]h=6.6\times 10^{-34}Js[/tex]

So energy of photon [tex]E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{121\times 10^{-12}}=1.636\times 10^{-15}J[/tex]

Mass of electron [tex]m=9.1\times 10^{-31}kg[/tex]

Velocity of electron is given [tex]v=56.9\times 10^6m/sec[/tex]

So kinetic energy of electron [tex]KE=\frac{1}{2}mv^2=\frac{1}{2}\times 9.1\times 10^{-31}\times (56.9\times 10^6)^2=1.473\times 10^{-15}J[/tex]

So binding energy = plank's energy - kinetic energy

[tex]=1.636\times 10^{-15}-1.473\times 10^{-15}=0.1633\times 10^{-15}J[/tex]

So binding energy will be [tex]0.1633\times 10^{-15}J[/tex]

The binding energy of an electron in an X-ray photoelectron experiment can be calculated by first determining the energy of the incoming photon and the kinetic energy of the ejected electron. The binding energy is then the difference between these two energies.

To calculate the binding energy of an electron in an X-ray photoelectron experiment, we'll require the principles of quantum mechanics and the concept of photoelectric effect. The energy of the incoming photon can be calculated using the equation:

E_photon = h*c/λ

where h is Planck's constant, c is the speed of light, and λ is the photon's wavelength. Converting the wavelength to the correct unit (meters), we get the energy of the photon.

Next, we calculate the kinetic energy of the ejected electron, using the equation:

E_kinetic = 0.5*m*v^2

where m is the mass of the electron, and v is its speed. Again, ensure that the speed is in the correct unit (meters per second).

The binding energy E_b of the electron then is the energy of the photon minus the kinetic energy of the electron:

E_b = E_photon - E_kinetic

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