The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?

Answer:

Explanation:

The speed of the astronaut can be found with the help of law of conservation of momentum .

mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of  astronaut.

Putting the values

77.5 x V = .94 x 12

V = .14554 m /s

This will be the uniform velocity of astronaut.

Distance to be covered = 37.3 m

time taken = distance / velocity

= 37.3 / .14554

= 256.28 s

= 4.27 minutes.

Answer:

Angular speed of rear wheel = 3.1425 rad/s

Explanation:

N1 = 16 rpm

w1 angular speed of big wheel = 2¶N/60 = (2 x 3.142 x 16)/60 = 1.676 rad/s

R1 = radius of big wheel = 15 cm = 15x10^-2 = 0.15 m

R2 = radius of small wheel = 8 cm = 0.08 m

w2 =?

Using w1R1 = w2R2

1.676 x 0.15 = w2 x 0.08

0.2514 = 0.08w2

w2 = 0.2514/0.08 = 3.1425 rad/s

Answer:

distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is  145.32 m

Explanation:

Given that :

The radius of the wheel  R =  28 cm

Initial angular speed of the wheel [tex]\theta_o = 16 rev/s[/tex]

= [tex]16*\frac{2}{1} *\frac{\pi}{1}rad/s[/tex]

= [tex]32 \pi\ rad/s[/tex]

= 100.5 rad/s

At time t = 9 s ;  the angle  rotated by wheel [tex]\theta_o = \omega_o t[/tex]

= 32 π × 9

= 905 rad

At time [tex]t_1 = 10 \ s[/tex]; the angular speed is definitely the same and the initial velocity [tex]\omega_1[/tex] is [tex]\omega_o = 32 \pi \ rads[/tex]

However ; after time  [tex]t_1 = 10 \ s[/tex] ; the angular acceleration of the wheel ∝ = 1.3 rad/s²

At time [tex]t_2 = 15\ s[/tex] ; angular speed of wheel [tex]\omega_2 = \omega_1 + \alpha \delta \ t[/tex]

[tex]\omega_2 = 32 \pi + 1.3( 15 - 10)[/tex]

[tex]\omega_2 =107 \ rad/s[/tex]

Now; the angle rotated by the wheel during time  [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is expressed as:

[tex]\theta_1 = \omega_1(t_2-t_1)+ \frac{1}{2} \alpha (t_2-t_1)^2\\\\\theta_1 = 32 \pi (15-10) + \frac{1}{2}*1.3*(15-10)^2\\\\\theta_1 = 519 \ rad[/tex]

From the question; we are being told that :

This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R.

i.e [tex]v_{cm} = R \omega[/tex]

then we can say :

acceleration [tex]a_{cm} = R \alpha[/tex]    and

linear distance [tex]s = R \theta[/tex]    indicating   how far did the center of the wheel move

So; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is ;

[tex]s = R \theta[/tex]

= 0.28 × 519

s = 145.32 m

Therefore; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is  145.32 m

The center of the wheel moved a distance of 0.418 meters between 10s and 15s.

In rolling motion without slipping, the instantaneous velocity of the atoms in contact with the ground is zero. This implies that the linear speed of the center of the wheel is equal to the product of the angular speed and the radius. To calculate the distance the center of the wheel moved between 10s and 15s, we need to find the average linear speed during that time period.

Therefore, the center of the wheel moved a distance of 0.418 meters between 10s and 15s.

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Answer:C

Explanation:

Radar waves will be useful to examine the surface of the planet.

Radio waves are the shorter wavelength microwaves ranging approximately 1 cm.

Short wavelength causes large reflection from large objects.  These waves are useful to detect smaller objects like water drops in the cloud.

They can penetrate clouds and that is why they are used in weather forecasting, air traffic control, etc.

The best wavelength of electromagnetic radiation to use when trying to examine a planet covered by a thick layer of clouds is radar waves, as they can penetrate the clouds and provide information about the surface beneath.

If you want to examine the surface of a planet that is completely covered by a thick layer of clouds all the time, the smartest wavelength of electromagnetic radiation to use would be C. Radar waves. The reason is that radar waves, which are a type of radio wave, can penetrate through clouds and provide information about the surface beneath. Other types of electromagnetic radiation such as visible light, x-rays, and ultra-violet would be absorbed or scattered by the clouds, preventing you from seeing the surface.

For example, radar is used here on Earth for weather prediction because it can 'see' through clouds, providing details about storm structures. Similarly, space probes such as the Magellan mission to Venus used radar wavelengths to map the surface of Venus, a planet notoriously covered in thick clouds.

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Answer:

1.62cm

Explanation:

a) object distance do = 25cm

image distance di = 1.71cm

we have the relation

1/f = 1/di +1/do

plugging in the values we get

f = 1.6cm

b) di = 1.71cm

do = 46cm

that gives

f = 1.65cm

C) do = 25cm

di =46 cm

plugging in the values in the above relation we get

f = 16.2cm

D) do = 36cm

f = 16.2cm

plugging in the equation

1/di = 1/f -1/do

plugging in the values we get

di = 29.4 cm

the new near point is 29.4cm

taking do = 29.4cm

di = 1.71 cm

we get focal length of eye as

f = 1.62cm

Final answer:

Calculating the focal length for presbyopia involves using the lens formula with different object distances. Accommodation factors in changes in near point due to age. Corrective lenses' focal lengths can also be found using similar optical principles.

Explanation:

When considering presbyopia, which is the age-related tendency to become far-sighted or hyperopic, there are several aspects of optics that come into play. To calculate the focal length of an eye's lens (f), we use the lens formula 1/f = 1/di + 1/do, where di is the image distance equal to the lens-to-retina distance, and do is the object distance. In this scenario, di remains at 1.71 cm, and the near point alterations represent changes in do.

A. When the object is at the near point of 25 cm, the focal length is found using the lens formula with do=25 cm.

B. As the near point shifts to 46 cm due to aging, we recalculate the focal length with this new do.

C. To correct presbyopia so the person can see an object at 25 cm, we need to determine the focal length of a corrective lens that compensates for the eye's inability to focus on close objects. This can be done by assuming that the corrective lens will bring the object effectively to the new near point and using the lens formula accordingly.

D. If the corrected vision allows seeing objects only if they are no closer than 36 cm, we can find the actual near point of the eye, which would dictate the adjusted focal length of the eye's lens needed to accommodate this near point.

Answer:

Explanation:

Since Ff, Fn, and Fg are the magnitudes of friction, normal, and gravitational forces on block 3, you can build the FBD (free bdy diagram) for block 3.

The FBD of the block 3 must include all the forces acting on the block 3.

The FBD is shown on the graph attached.

Observe this:

Answer:

the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂

Explanation:

The center of mass, the point of a system where all external forces can be applied, is defined by

        [tex]x_{cm}[/tex] = 1 /M ∑ [tex]x_{i} m_{i}[/tex]

where M is the total mass of the system, x_{i} m_{i} are the position and mass of each item in the system,

let's apply this equation to our case

the total mass is

    M = m₁ + m₂

for the calculations we must fix a reference system, we will place it in the second mass (m₂)

     x_{cm}= 1/M (m₁ d + m₂ 0)

where d is the distance between the two masses in this case d = 10 m

    x_{cm} = m₁ / (m₁ + m₂) d

    x_{cm} = m₁ / (m₁ + m₂) 10

as the mass m₁ <m₂

Let us analyze the answer if the masses sides were equal the center of mass would be x_{cm} = 5 m, but since m₁ <m₂ the center of mass must be closer to m₂.

Therefore the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂

Answer:

For each 1.5 cm the input piston moves, then output piston will move 0.5 cm

Explanation:

Let cross-sectional area of the input piston = A

Let cross-sectional area of the output piston = 3A

pressure (P) = Force (F) / Area (A)

F = PA

For a constant gravitational force on the inlet and outlet piston;

P₁A₁ = P₂A₂

But pressure = ρgh

where;

ρ is density of water

g is acceleration due to gravity

h is the distance or height moved by the piston

(ρg)h₁A₁ = (ρg)h₂A₂

h₁A₁ = h₂A₂

Area of output piston = 3 times area of input piston

h₁A₁ = h₂(3A₁)

For each 1.5 cm the input piston moves, then output piston will move;

1.5A₁ =  h₂(3A₁)

1.5 = 3h₂

h₂ = 1.5 / 3

h₂ = 0.5 cm

Thus, for each 1.5 cm the input piston moves, then output piston will move 0.5 cm

Answer:

Student b

Explanation:

Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)

This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.

Answer:

Student B is correct

Explanation:

The battery acts like a pump to drive the electric charge. The electric charge comes from the conducting material but the battery provides a potential difference or a gradient of flow for this charge which is necessary for the movement of the charge along the conductor.

Answer:

Possible options:

A. Moment of inertia

B. Angular acceleration

C. Center of mass

D. Torque

Answer is B. Angular acceleration

Explanation:

Angular acceleration is the time rate of change of angular velocity. In three dimensions, it is a pseudovector. In SI units, it is measured in radians per second squared (rad/s2) and is usually denoted by the Greek letter alpha (α).

The angular acceleration should be with respect to the shorter pencil and it should be larger.

Angular acceleration refers to the time that shows the rate of change with respect to the angular velocity. In terms of three dimensions, it represents the pseudovector. Also, in SI units, it should be measured in radians per second squared and should be denoted by the  Greek letter alpha (α).

Therefore, The angular acceleration should be with respect to the shorter pencil and it should be larger.

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Answer:

At the edge, angular acceleration = 3.025 rad/s2

At halfway = 13.11 rad/s2

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

263.8 m/s2

Explanation:

Assume this is a solid disk, we can find its moments of inertia:

[tex]I = mr^2/2 = 3.8*1.6^2/2 = 4.864 kgm^2[/tex]

The torque T generated by force F = 18.4N is:

[tex]T = Fr = 18.4*1.6 = 29.44 Nm[/tex]

So the angular acceleration of the disk according to Newton's 2nd law is:

[tex]\alpha = T/I = 29.44 / 18.4 = 6.05 rad/s^2[/tex]

If the disk starts from rest, then after 3s its angular speed is

[tex]\omega = \alpha \Delta t = 6.05*3 = 18.16 rad/s[/tex]

And so its radial acceleration at this time and half way from the center to the edge is:

[tex]a_r = \omega^2(r/2) = 18.16^2*(1.6/2) = 263.8 m/s^2[/tex]

Note that this value is the same anywhere

Answer:

155 N

Explanation:

We are given following data for a uniform, solid, horizontal disk:  

r = 1.5 m

m= 110 kg

t=2 s

w = 0.6  

Torque is given by:  

т = F*r

  =I*α

Solving it for the force exerted on the rope:  

F = I*α/r

  = (1/2*m*r^2)*(2π*w/t )/r

  = (1/2*110*1.5^2)*(2π*0.6/2 )/1.5

  = 155 N

Answer:

λ = 4.97 x 10⁻⁷ m = 497 nm

Explanation:

The formula for the distance of a bright fringe from center in Young's double slit experiment, is given by:

y = mλL/d

where,

y = distance of bright fring from center = 3.22 mm = 3.22 x 10⁻³ m

m = No. of bright fringe = 1

L = Distance between slits and screen = 3.24 m

d = slit separation = 0.5 mm = 0.5 x 10⁻³ m

λ = wavelength of light = ?

3.22 x 10⁻³ m = (1)(λ)(3.24 m)/(0.5 x 10⁻³ m)

(3.22 x 10⁻³ m)(0.5 x 10⁻³ m)/(3.24 m) = λ

λ = 4.97 x 10⁻⁷ m = 497 nm

Answer:

Nuclear engineer

Explanation:

Because it requires a lot of movements during that career.

Answer:

the ratio of the deuteron's orbital radius to the proton's orbital radius is 2 : 1

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

See attached file pls

Explanation:

Answer:

Let's analyse the definition and applications of angular momentum, and its relation with torque.

First of all, it's important to consider that the angular momentum is a property of rotational dynamics. Also, it's the analogue of the linear momentum.

Mathematically, the angular momentum is defined as

[tex]L=r \times p[/tex]

Where [tex]L[/tex] represents the angular momentum vector, [tex]r[/tex] represents is the position vector and [tex]p[/tex] is the linear momentum vector.

Notice that the angular momentum is also a vector, which is the cross product  of two vectorial magnitudes. In other words, the direction of the resulting vector (linear momentum) follows the right hand rule, which means that the resulting direction is according to the rotation direction, also means that the cross product is not commutative, which is a common assumption students make.

Now, the realtion between angular momentum and torque is that the change of the angular momentum with respect to time is equivalent to its torque:

[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] =\tau[/tex]

Remember that torque is defined as [tex]\sum \tau = r \times \sum F[/tex], and the derivative of the cross product is

[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] = \frac{dr}{dt} \times p + r \times \frac{dp}{dt}[/tex]

Then,

[tex]\frac{d}{dt}[L] =(v \times mv)+r \times \frac{dp}{dt}[/tex]

But, [tex]v \times mv = 0[/tex], because those vector are parallel.

So, [tex]\frac{d}{dt}[L] = r\times \sum F = \tau[/tex]

At this point, we demonstrate it the relation between torque and rotational momentum.

In words, the net torque on a particle is equal to the rate of change of the angular momentum with respect to time.

Now, the application of angular momentum can be seen in skating spins, notice that when the skater puts his arms closer to its body, he'll rotate faster. The reason of this phenomenon is because arms represents rotating mass and the axis is the body, so the postion of this arm mass changes to zero distance to the rotational axis, that will increase the angular momentum, making higher. If the angular momentum is higher, the torque will be also higher, that's way the skater increses its rotational velocity.

Angular momentum is the rotational equivalent of linear momentum, defined as L = Iω for extended bodies. It is conserved when net external torque is zero, similar to how linear momentum is conserved in the absence of external forces. Torque and angular momentum are related by net τ = ΔL/Δt.

In physics, the concept of angular momentum is the rotational equivalent of linear momentum. It is defined as L = r × p for a single particle, where r is the position vector and p is the linear momentum vector. For an extended body rotating about an axis of symmetry, angular momentum (L) is given by the product of the moment of inertia (I) and angular velocity (ω): L = Iω.

Just as linear momentum is conserved in the absence of external forces, angular momentum is conserved when the net external torque is zero. This principle is crucial in various applications, such as planetary motion and rotating machinery.

The relationship between torque (τ) and angular momentum is given by the equation: net τ = ΔL/Δt, meaning the net torque acting on a system is equal to the rate of change of angular momentum.

Answer:

the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

Explanation:

Using Rayleigh criterion for the limiting angle of resolution of an eye

[tex]\theta = \frac{1.22\lambda }{D } \\ \\ \theta = \frac{1.22*506 *10^{-9} }{4.90*10^{-3}m}[/tex]

[tex]\theta = 1.2598*10^{-4}[/tex] rad

[tex]\theta = 125.98*10^{-6} \ rad[/tex]

Thus; the separation  between the two sources is expressed as:

[tex]\theta = \frac{y}{L} \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^{-6} \ rad) \\ \\ y = 31.495 \ m[/tex]

Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

Final answer:

The approximate size of the smallest object that astronauts can resolve by eye when orbiting 250 km above the Earth is approximately 628 meters.

Explanation:

The approximate size of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth can be determined using the Rayleigh criterion. According to the Rayleigh criterion, the resolution of the eye is determined by the angle of resolution, which is given by:

angle of resolution = 1.22 * (wavelength / pupil diameter)

Using the given values, we can calculate the angle of resolution:

angle of resolution = 1.22 * (506 nm / 4.90 mm) = 1.2611957657 x 10^-3 radians

To determine the approximate size of the smallest object, we need to find the linear size corresponding to this angular resolution at a distance of 250 km:

linear size = 2 * distance * tan(angle of resolution)

linear size = 2 * 250000 m * tan(1.2611957657 x 10^-3 radians) = 628.1079000855 m

Therefore, the approximate size of the smallest object that astronauts can resolve by eye when they are orbiting 250 km above the Earth is approximately 628 meters.

Answer:

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle, [tex]q=6.41\ \mu C=6.41\times 10^{-6}\ C[/tex]

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force, [tex]F=6.4\times 10^{-3}\ N[/tex]

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

[tex]F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s[/tex]

So, the particle's velocity is 212.15 m/s.

Answer:

Explanation:

Find attached the solution

Answer:

Explanation:

angular acceleration α = 5 rad /s ²

θ = 1/2 α t² ,     θ is angle of rotation , t is time .

= .5 x 5 x 4²

= 40 rad .

θ = l / r , θ is angle formed , l is length unwound , r is radius o wheel .

l = θ x r

= 40 x .1

= 4 m .

Answer:

A) X(0.5) = 0 ft

B) X(0.75) = 0.023 ft.

Explanation: Given that the result is measured in feet.

x(t) = −e−4t + 2te−4t

Factorizing the above equation lead to

x(t) = e^-4t( -1 + 2t )

The mass passes through the equilibrium position when x(t) = 0

0 = e^-4t( -1 + 2t )

-1 + 2t = 0

2t = 1

t = 0.5s

x(t) = e^-4t( -1 + 2t )

Substitute t = 0.5

x(0.5) = e^-4(0.5)(-1 + 2(0.5))

X(0.5) = 0

Also given that mass attains its extreme displacement from the equilibrium position is t = 3/ 4 

= 0.75 seconds

x(t) = e^-4t( -1 + 2t )

x(t) = e^-4(0.75)( -1 + 2(0.75) )

X(0.75) = (e^-3)(0.5)

X(0.75) = 0.023 ft.

Answer:

(A) Acceleration will be [tex]1.145rad/sec^2[/tex]

(B) Coefficient of static friction will be 0.116

Explanation:

We have given angular speed

[tex]\omega =33rpm=\frac{2\pi \times 33}{60}=3.454rad/sec[/tex]

Distance from the axis r = 9.6 cm = 0.096 m

(a) Acceleration is equal to

[tex]a_c=\omega ^2r=3.454^2\times 0.096=1.145rad/sec^2[/tex]

(b) For seed is not to slip

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

[tex]\mu =\frac{a}{g}[/tex]

[tex]\mu =\frac{1.145}{9.8}=0.116[/tex]

So coefficient of static friction will be 0.116

Answer:

A

Explanation:

Because in block friction does not work hence case of cylinder has more total K.E

Complete Question

A thin film of cooking oil (n=1.43) is spread on a puddle of water (n=1.34). What is the minimum thickness Dmin of the oil that will strongly reflect blue light having a wavelength in air of 451 nm, at normal incidence? Dmin= nm What are the next three thicknesses that will also strongly reflect blue light of the same wavelength, at normal incidence?

a)   158 nm, 237 nm, and 315 nm

b)   237 nm, 394 nm, and 552 nm

c)   473 nm, 788 nm, and 1100 nm

d)  361 nm, 602 nm, and 842 nm

e)  315 nm, 473 nm, and 631 nm

Answer:

The minimum thickness is   [tex]t= 78.8nm[/tex]

The correct option is B

Explanation:

  From the question we are told that

      The refractive index of cooking oil is [tex]n _c = 1.43[/tex]

       The refractive index of water is [tex]n_w = 1.34[/tex]

      The  wavelength of reflection is is  [tex]\lambda _ r = 451nm[/tex]

The formula for the thickness of the oil film is mathematically represented as

        [tex]2 n t = (m + \frac{1}{2} ) \lambda[/tex]

Where n is the refractive index of oil

            m is the integer number of fringe

            t is the thickness

   for a minimum reflection  m= 0

Now making t the subject of the formula

         [tex]t = \frac{(m + \frac{1}{2} \lambda ) }{2 n}[/tex]

Substituting value

        [tex]t = \frac{(0 + 0.5) * 451 *10^{-9}}{2 * 1.43}[/tex]

          [tex]t= 78.8nm[/tex]

For the next thickness m = 1

   so we have

        [tex]t_1 = \frac{(1 + 0.5 ) * 481}{2 * 1.43}[/tex]

            [tex]= 237nm[/tex]

For the next thickness m = 2

     so we have that  

        [tex]t_2 = \frac{(2 +0.5) *451 *10^{-9}}{2 * 1.43}[/tex]

             [tex]= 394nm[/tex]

For the next thickness m = 3

     so we have that

         [tex]t_2 = \frac{(3 +0.5) *451 *10^{-9}}{2 * 1.43}[/tex]

             [tex]= 552nm[/tex]

According to the question,

(a)

→              [tex]2nt = (m+\frac{1}{2} ) \lambda[/tex]

By substituting the values, we get

→ [tex]2\times 1.43\times t = (0+\frac{1}{2} )481[/tex]

→                  [tex]t = 84.1 \ nm[/tex]

(b)

The next 3 thicknesses are:

→ [tex]2\times 1.43\times t = (1+\frac{1}{2} ) 481[/tex]

                   [tex]t = 252 \ nm[/tex]

→ [tex]2\times 1.43\times t = (2+\frac{1}{2} )481[/tex]

                   [tex]t = 420 \ nm[/tex]

→ [tex]2\times 43\times t = (3+\frac{1}{2} ) 481[/tex]

                [tex]t = 588 \ nm[/tex]

Thus the above approach is correct.

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Answer:

Explanation:

We shall consider tension to be T.

For motion of bucket in downward direction

mg - T = ma , m is mass of bucket , a is linear acceleration of bucket

for rotational motion of wheel

Tortque by force Tr = I x α        I is moment of inertia ,   α  is angular acceleration , r is radius of the wheel.

Putting the values in the equation above

mg -  I x α / r  = ma

mg -  1/2 M r ²x a / r²   = ma

mg - 1/2 M a = ma

mg = 1/2 M a + ma

a = mg / (m + .5 M)

= 16 X 9.8 / (16 +.5 x 11.4)

= 7.22 m /s²

mg - T = ma

T = m( g - a )

= 16 x ( 9.8 - 7.22 )

= 41.28 N.

b )

v² = u² + 2 a h , v is velocity when bucket falls by height h

= 0 + 2 x 7.22 x 10.1

= 145.84

v = 12.07

c )

v = u + at , t is time of fall

12.07 = 0 + 7.22 t

t = 1.67 s

d )

Force by axle on cylinder = T

= 41.28 N .

By using the principle of energy conservation and accounting for the charges and masses of the proton and alpha particle, we can establish that the speed of the proton when the distance between the particles doubles is given by the square root of four times Coulomb's constant times the charge of the proton squared, divided by the mass of the proton and the original distance.

To solve the problem, we can apply the principle of energy conservation. In the initial state, the system has potential energy, which is due to the electrostatic interactions between the proton and the alpha particle. When the particles begin to move apart, some of this potential energy is converted into kinetic energy. Given that both particles are positively charged, the potential energy U of the interaction is given by Coulomb's law:

U = k * Q1 * Q2 / r

where Q1 and Q2 are the charges of the particles, r is the distance between them and k = 8.99 * 10^9 N * m^2/C^2 is Coulomb's constant. The alpha particle has 2e, and the proton has e, leading to the potential energy of the system being: U_initial = 2 * k * e^2 / r.

When the distance between the proton and the alpha particle doubles, the potential energy becomes: U_final = k * e^2 / (2r).

The change in potential energy (delta U) thus equals the kinetic energy of the proton (as the alpha particle is more massive, its kinetic energy can be neglected): delta U = U_initial - U_final = K_proton = (1/2) * m * v^2, where m and v are the mass and speed of the proton, respectively. Solving this equation for v, we get that the speed of the proton is: v = sqrt((4 * k * e^2)/(m * r)).

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Answer:

The approximate number of decays  this represent  is  [tex]N= 23*10^{10}[/tex]  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is [tex]I_a = 2.28 \ mSv[/tex]

     The source of the radiation is [tex]S = 5.49 MeV \ alpha \ particle[/tex]

 Generally

            [tex]1 \ J/kg = 1000 mSv[/tex]

   Therefore  [tex]2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg[/tex]

Also  [tex]1eV = 1.602 *10^{-19}J[/tex]

  Therefore  [tex]2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg[/tex]

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        [tex]1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x[/tex]

Cross-multiplying and making x the subject

           [tex]x = 88.7 * 1.423*10^{16} eV[/tex]

              [tex]x = 126.2*10^{16}eV[/tex]

Therefore the total  energy  deposited is [tex]x = 126.2*10^{16}eV[/tex]

The approximate number of decays  this represent  is mathematically evaluated as

            N = [tex]\frac{x}{S}[/tex]

Where n is the approximate number of decay

   Substituting values

                             [tex]N = \frac{126 .2*10^{16}}{5.49*10^6}[/tex]  

                                  [tex]N= 23*10^{10}[/tex]  

The energy deposited in the body from radon exposure and the number of decays it represents. It also discusses the significance of radon-222 as a source of radiation exposure.

The energy deposited in your body each year from radon can be calculated by converting the dose equivalent to energy. Given that the average American receives 2.28 mSv annually and all of it comes from a 5.49 MeV alpha particle emitted by Rn-222, the energy deposited would be 5.49 MeV.

To determine how many decays this represents, you divide the total energy deposited by the energy per decay. In this case, 5.49 MeV divided by 5.49 MeV per decay gives you 1 decay annually.

Radon-222 is a significant source of radiation exposure due to its radioactivity and ability to seep into homes from the ground. It highlights the importance of understanding the impact of radioactive elements on our environment.

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

The gas pressure inside the box is [tex]\( 1.018 \times 10^5 \, \text{Pa} \)[/tex] to three significant figures.

To find the gas pressure [tex]\( p_{\text{gas}} \)[/tex] inside the box using the information provided, we'll follow the hydrostatic equilibrium principles as outlined.

The density of mercury [tex](\( \rho \)) = \( 1.36 \times 10^4 \, \text{kg/m}^3 \)[/tex]

The height of the mercury column in the left arm [tex](\( h_1 \)) = 10.0 cm = 0.10 m[/tex]

The height of the mercury column in the right arm [tex](\( h_2 \)) = 6.00 cm = 0.06 m[/tex]

Atmospheric pressure [tex](\( p_{\text{atm}} \)) = 1.00 atm = \( 1.013 \times 10^5 \, \text{Pa} \)[/tex]

Acceleration due to gravity [tex](\( g \)) = \( 9.81 \, \text{m/s}^2 \)[/tex]

1. Convert the heights to meters:

[tex]\[ h_1 = 0.10 \, \text{m} \] \[ h_2 = 0.06 \, \text{m} \][/tex]

2. Draw a schematic:

The left arm of the U-tube is connected to the box containing the gas. The right arm is open to the atmosphere. Mercury levels in the U-tube indicate that the pressure on the left side (due to the gas) is higher than the pressure on the right side (atmospheric pressure).

3. Pressure at the interface between mercury and gas (left side):

[tex]\[ p_{\text{left}} = p_{\text{gas}} \][/tex]

4. Pressure at the interface on the open side (right side):

[tex]\[ p_{\text{right}} = p_{\text{atm}} \][/tex]

5. Difference in mercury column height:

[tex]\[ \Delta h = h_1 - h_2 = 0.10 \, \text{m} - 0.06 \, \text{m} = 0.04 \, \text{m} \][/tex]

6. Hydrostatic pressure difference:

  The difference in height of the mercury columns corresponds to the pressure difference due to the gas in the box and the atmospheric pressure:

[tex]\[ \Delta p = \rho g \Delta h \] \[ \Delta p = (1.36 \times 10^4 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (0.04 \, \text{m}) \] \[ \Delta p = 1.36 \times 10^4 \times 9.81 \times 0.04 \] \[ \Delta p = 532.224 \, \text{Pa} \][/tex]

7. Calculate the gas pressure: Since the pressure in the gas must account for the atmospheric pressure plus the pressure due to the height difference of the mercury:

[tex]\[ p_{\text{gas}} = p_{\text{atm}} + \Delta p \] \[ p_{\text{gas}} = 1.013 \times 10^5 \, \text{Pa} + 532.224 \, \text{Pa} \] \[ p_{\text{gas}} = 1.01832 \times 10^5 \, \text{Pa} \][/tex]

[tex]\[p_{\text{gas}} = 1.018 \times 10^5 \, \text{Pa}\][/tex]

Complete question:- To practice Tactics Box 13.1 Hydrostatics.

In problems about liquids in hydrostatic equilibrium, you often need to find the pressure at some point in the liquid. This Tactics Box outlines a set of rules for thinking about such hydrostatic problems.

Answer:

Assume patmos=1.00atmpatmos=1.00atm. What is the gas pressure?

Express your answer in pascals to three significant figures.

Pgas=_____ Pa

EDIT (Needed Information):

Draw a picture. Show open surfaces, pistons, boundaries, and other features that affect pressure. Include height and area measurements and fluid densities. Identify the points at which you need to find the pressure. These objects make up the system; the environment is everything else.

Determine the pressure p0p0 at surfaces.

Surface open to the air: p0=patmosp0=patmos, usually 1 atmatm.

Surface in contact with a gas: p0=pgasp0=pgas.

Closed surface: p0=F/Ap0=F/A, where FFF is the force the surface, such as a piston, exerts on the fluid and AAA is the area of the surface.

Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point along a horizontal line.

Allow for gauge pressure. Pressure gauges read pg=p−1atmpg=p−1atm.

Use the hydrostatic pressure equation: p=p0+ρghp=p0+ρgh, where ρρrho is the density of the fluid, ggg is the acceleration due to gravity, and hhh is the height of the fluid.

Use these rules to work out the following problem: A U-shaped tube is connected to a box at one end and open to the air at the other end. The box is full of gas at pressure pgaspgasp_gas, and the tube is filled with mercury of density 1.36×104 kg/m3kg/m3 . When the liquid in the tube reaches static equilibrium, the mercury column is h1h1h_1 = 10.0 cmcm high in the left arm and h2h2h_2 = 6.00 cmcm high in the right arm, as shown in the figure.(Figure 1) What is the gas pressure pgaspgasp_gas inside the box?

The moment of inertia of the given cube, which is the rotational equivalent of mass in linear motion, is 0.000075 kg.m^2. This is calculated by applying the formula I = (M*s^2)/6 with given parameters. The parallel axis theorem helps in finding the moment of inertia about any axis parallel to and a distance away from an object's center of mass.

The moment of inertia of a cube about an axis normal to and through the center of one of its faces is calculated using the formula I = (M*s^2)/6, where M is mass, and s is the side length of the cube. In this case, M = 0.500 kg and s = 0.030 m. Therefore, the moment of inertia I = (0.500 kg * (0.030 m)^2)/6 = 0.000075 kg.m^2. Moment of inertia can be thought of as the rotational equivalent of mass for linear motion.

The moment of inertia of a cube, or any object, about any axis through its center of mass can be calculated using the parallel axis theorem. The theorem links the moment of inertia about an axis passing through the object's center of mass to the moment of inertia about any other parallel axis. It states that the moment of inertia about any axis parallel to and a distance d away from the axis through the center of mass is equal to the moment of inertia of the object about the axis through the center of mass plus the mass of the object times the square of the distance between the axes (I=I_cm + md^2). Therefore, the moment of inertia is also related to the distribution of mass within an object.

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