Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

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Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

Answer:

The electric field is [tex]E = 5.25 V/m[/tex]

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]

Generally the peak magnitude of the electric field is mathematically represented as

         [tex]E = c * B[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

       [tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]

       [tex]E = 5.25 V/m[/tex]

The peak magnitude of the electric field will be "5.25 V/m".

According to the question,

Magnetic field's peak magnitude, B = 17.5 nT or,

                                                           = 17.5 × 10⁻⁹ T

Speed of light, c = 3.0 × 10⁸ m/s

We know the relation,

→ E = c × B

By substituting the values, we get

      = 3.0 × 10⁸ × 17.5 × 10⁻⁹

      = 5.25 V/m

Thus the above approach is appropriate.

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Answer:

a. to the west.

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer:

Vrel= 0.75c

Explanation:

See attached file

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Hope this helps you

Answer:

The wavelength is  [tex]\lambda = 589 nm[/tex]

Explanation:

From the question we are told that

    The  distance of the mirror shift  is  [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]

      The number of fringe shift is  n =  792

Generally the wavelength producing this fringes is mathematically represented as

               [tex]\lambda = \frac{ 2 * k }{ n }[/tex]

substituting values

              [tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]

             [tex]\lambda = 5.885 *10^{-7} \ m[/tex]

            [tex]\lambda = 589 nm[/tex]

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

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Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet

By applying the integrations,

(a) [tex]S = 96t-16t^2+432[/tex]

(b) Time will be "t = 9".

(c) Height will be "576"

Given:

Height,

Initial velocity,

According to the question,

(a)

Integrate v:

Initial Condition,

→ [tex]S = 96t-16t^2+432[/tex]

(b)

Hits the ground when,

→ [tex]0=96t-16t^2+432[/tex]

→ [tex]t =9[/tex]

(c)

Maximum height when,

→ [tex]0 = 96-32 t[/tex]

→ [tex]t = 3[/tex]

Now,

→ [tex]S = 96\times 3-16\times 3^2+432[/tex]

      [tex]= 576[/tex]

Thus the answer above is correct.

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Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

Answer:

The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]

    The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]

    The  number of slit is  N =  1000 slits

    The order of the maxima is  m =  2

Generally the spacing between the slit is mathematically represented as

         [tex]d = \frac{k}{N}[/tex]

substituting values

        [tex]d = \frac{ 0.01}{1000}[/tex]

       [tex]d = 1.0 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is

       [tex]d\ sin(\theta ) = m * \lambda[/tex]

substituting values

      [tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]

       [tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]

      [tex]\theta = 6.315^o[/tex]

Generally the dispersion is mathematically represented as

           [tex]D = \frac{ m }{d cos(\theta )}[/tex]

substituting values

          [tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]

           [tex]D = 2.01220 *10^{5} \ rad/m[/tex]

Answer:

"Endoscope" is the correct answer.

Explanation:

Answer:

The energy of one of its photons is 1.391 x 10⁻¹⁸ J

Explanation:

Given;

wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m

The energy of one photon of the UVC light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency of the light

f = c / λ

where;

c is speed of light = 3 x 10⁸ m/s

λ  is wavelength

substitute in the value of f into the main equation;

E = hf

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]

Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J

Answer and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]

Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]

At rest ;

their velocities will be zero, i.e

[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0

The initial momentum for this process can be represented as :

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0

after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]

The we can say their final momentum is:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] +  [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] =  [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Since the initial velocities are stating at rest then ; u = 0

[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] +   [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]  = 0

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So [tex]m_{Ri} > m_{Pa}[/tex] ;

Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] =  [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]

The ratio is

[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]

[tex]v_{Pa} >v_{Ri}[/tex]

Therefore, Paula will have a greater speed than Ricardo after the push-off.

(A) Both the skaters have the same magnitude of momentum.

(B) Paula has greater speed after push-off.

Given that two skaters Paula and Ricardo are initially at rest.

Ricardo weighs more than Paula.

Let us assume that the mass of Ricardo is M, and the mass of Paula is m.

Let their final velocities be V and v respectively.

(A) Initially, both are at rest.

So the initial momentum of Paula and Ricardo is zero.

According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.

Initial momentum = final momentum

0 = MV + mv

MV = -mv

So, both of them have the same magnitude of momentum, but in opposite directions.

(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:

MV = mv

M/m = v/V

Now, we know that M>m

so, M/m > 1

therefore:

v/V > 1

v > V

So, Paula has greater speed.

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Answer:

380 kHz

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

where v is the speed of a wave

λ is the wavelength of the wave

f is the frequency of the wave

f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ 380 kHz

Answer:

Check your router connections then restart your router.

Explanation:

Answer:

Check your router connections then restart your router.

Explanation:

Most internet access comes from routers so the problem is most likely the router.

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

Answer:

[tex]v=1.24\times 10^8\ m/s[/tex]

Explanation:

Given that,

The refractive index of benzene is 2.419

We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,

[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]

So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m