A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to move this charge 50 cm along a path making an angle of 40

Answer:

a

  [tex]B = 0.0533 \ T[/tex]

b

  [tex]B = 0.04 \ T[/tex]

Explanation:

From the question we are told that

   The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]

   The  outer radius is  [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]

    The nu umber of turns is  [tex]N = 960[/tex]

    The current it is carrying is  [tex]I = 2. 5 A[/tex]

Generally the magnetic field is mathematically represented as

      [tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with a constant value    

            [tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]

And the given distance where the magnetic field is felt is  r =  0.9 cm  =  0.009 m

Now  substituting values

     [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]

    [tex]B = 0.0533 \ T[/tex]

    Fro the second question the distance of the position considered from the center is  r =  1.2 cm  =  0.012 m

So the  magnetic field is  

        [tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]

        [tex]B = 0.04 \ T[/tex]

The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.

The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.

The given parameters;

The magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]

The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;

[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]

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Answer:

momentum of the coupled cars V =  1.77 m/s

kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J

Explanation:

given

m₁ =3.05 × 10⁴kg

u₁ =2.90m/s

m₂=6.10× 10⁴kg

u₂=1.20m/s

using law of conservation of momentum

m₁u₁ + m₂u₂ = (m₁ + m₂) V

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V

V =  1.617×10⁵/9.15×10⁴

V = 1.77m/s

K.E =1/2mV²

ΔK.E = K.E(final) - K.E(initial)

ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔK.E = -2.892×10⁴J

The final speed is 1.77 m/s

The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]

2.892×10⁴J of energy is converted.

Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.

Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg

The initial speed of the railroad boxcar is u₁ = 2.90m/s

Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg

And the initial speed be u₂ = 1.20m/s

The initial momentum of the first car is:

m₁u₁ = 3.05 × 10⁴ × 2.90 =  8.84 × 10⁴ kgm/s

The initial momentum of the coupled car is:

m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s

Let the final speed after all the boxcars are coupled be v

From the law of conservation of momentum, we get:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv

v =  1.617×10⁵/9.15×10⁴

v = 1.77m/s

The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:

ΔKE = K.E(final) - K.E(initial)

ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² -  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴

ΔKE = -2.892×10⁴J

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Answer:

The magnitude of the momentum is 49145.19 kg.m/s

The direction of the two vehicles is 44.6° North West

Explanation:

Given;

speed of first vehicle, v₁ = 14 m/s (East to west)

mass of first vehicle, m₁ = 1500 kg

speed of second vehicle, v₂ = 23 m/s (South to North)

momentum of the first vehicle in x-direction (E to W is in negative x-direction)

[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]

momentum of the second vehicle in y-direction (S to N is in positive y-direction)

[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]

Magnitude of the momentum of the system;

[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]

Direction of the two vehicles;

[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West

Answer:

c. tension, gravity, and the centripetal force

Explanation:

The ball experiences a variety of force as explained below.

Gravity force acts on the body due to its mass and the acceleration due to gravity. The gravity force on every object on earth due to its mass keeps all object on the surface of the earth.

Although the car moves around in circle, centripetal towards the center of the radius of turn exists on the ball. This centripetal force is due to the constantly changing direction of the circular motion, resulting in a force away from the center. The centripetal force keeps the ball from swinging off away from the center of turn.

Tension force on the string holds the ball against falling towards the earth under its own weight, and also from swinging away from the center of turn of the car. Tension force holds the ball relatively fixed in its vertical position in the car.

Answer:

18.75 rad/s

Explanation:

Moment of inertia of the disk;

I_d = ½ × m_disk × r²

I_d = ½ × 10 × 4²

I_d = 80 kg.m²

I_package = m_pack × r²

Now,it's at 2m from the centre, thus;

I_package = 5 × 2²

I_package = 20 Kg.m²

From conservation of momentum;

(I_disk + I_package)ω1 = I_disk × ω2

Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.

Thus;

Plugging in the relevant values, we obtain;

(80 + 20)15 = 80 × ω2

1500 = 80ω2

ω2 = 1500/80

ω2 = 18.75 rad/s

Answer:

Because closed magnetic field loops have to be formed between both ends of the magnet, a magnet will always have two poles.

Explanation:

Magnetic Monopoles do not exist in nature because a magnetic field always forms a loop that runs from one end of the magnet to the other.

Since this loop of the magnetic field has an origination and termination point which are at the two ends of the magnet (North and South poles).  A magnet will always be bipolar which is in this case, North and South; even at an atomic level.

Answer:

[tex]Q'=\sqrt{6}Q[/tex]

Explanation:

You have that a parallel plate capacitor has a total energy of E when the distance between the plates is d and the charge on each plate is Q.

You take into account the following formula for the stored energy in the capacitor:

[tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex]          (1)

The capacitance C of the parallel plate capacitor is given by the following formula is:

[tex]C=\epsilon_o\frac{A}{d}[/tex]          (2)

A: area of the plates

ε0: dielectric permittivity of vacuum

You replace the expression (2) into the equation (1):

[tex]E=\frac{1}{2}\frac{Q^2A}{\epsilon_o d}[/tex]       (3)

the previous formula is the expression for the total energy stored for the given parameters A, d and Q.

If the distance between the plates is twice and it is required that the energy is three times the initial energy, to find the value of the charge you use the equation (3):

[tex]E'=\frac{1}{2}\frac{Q'^2A}{\epsilon_o d'}[/tex]        (4)

d' = 2d

E' = 3E

Q': required charge

You replace the values of d' and E' in the equation (4) and then divide the result with the equation (3):

[tex]3E=\frac{1}{2}\frac{Q'^2A}{\epsilon_o(2d)}=\frac{1}{4}\frac{Q'^2A}{\epsilon_od}\\\\\frac{3E}{E}=\frac{1/4\frac{Q'^2A}{\epsilon_od}}{1/2\frac{Q^2A}{\epsilon_o d}}\\\\3=\frac{1}{2}\frac{Q'^2}{Q^2}[/tex]

Finally, you solve for Q':

[tex]3=\frac{1}{2}\frac{Q'^2}{Q^2}\\\\Q'=\sqrt{6}Q[/tex]

Then, the required charge is √6Q , to obtain three times the initial energy E, when the distance between plates is doubled.

Answer:

[tex]v=21.36\,\,\frac{m}{s}\\[/tex]

[tex]m=1.2357\,\,kg[/tex]

Explanation:

Recall the formula for linear momentum (p):

[tex]p = m\,v[/tex]  which in our case equals 26.4 kg m/s

and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:

[tex]K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}[/tex]

Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:

[tex]K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg[/tex]

Now by knowing the particle's mass, we use the momentum formula to find its speed:

[tex]p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}[/tex]

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex]    ..............1

put here value and we get

F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]

F(r) =  11.45 N

so magnitude of the resultant of forces is 11.45 N

Complete Question:

Given [tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex] at a point. What is the force per unit area at this point acting normal to the surface with[tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex]   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, [tex]\sigma_n = 28 MPa[/tex]

There are shear stresses acting on the surface since [tex]\tau \neq 0[/tex]

Explanation:

[tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex]

equation of the normal, [tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex]

[tex]\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]

Traction vector on n, [tex]T_n = \sigma \b n[/tex]

[tex]T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]

[tex]T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right][/tex]

[tex]T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z[/tex]

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

[tex]\sigma_n = T_n . \b n[/tex]

[tex]\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa[/tex]

If the shear stress, [tex]\tau[/tex], is calculated and it is not equal to zero, this means there are shear stresses.

[tex]\tau = T_n - \sigma_n \b n[/tex]

[tex]\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z[/tex]

[tex]\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa[/tex]

Since [tex]\tau \neq 0[/tex], there are shear stresses acting on the surface.

Answer:

Drop to half of the previous value

Explanation:

Energy stored in capacitor is inversly propotional to the distance between the plates.

If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor drops to half its previous value.

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is  given by

C = ε₀ A/d

where, ε₀ is the permittivity of free space, A = area of cross section of plates and d is the distance between them.

Capacitance is inversely proportional to the distance between them. So, if distance is doubled, the capacitance decreases to half its original value.

Thus, the correct option is 8.

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Question:

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.

Answer:

S = 5.508 × 10¹¹m

V = 2.62 × 10⁸ m/s

Explanation:

The radius of the orbit of Jupiter, Rj is 43.2 light-minutes

radius of the orbit of Mars, Rm is 12.6 light-minutes

Distance travelled S = (Rj - Rm)

= 43.2 - 12.6 = 30.6 light- minutes

= 30.6 × (3 ×10⁸m/s) × 60 s

= 5.508 × 10¹¹m

time = 35mins = (35 × 60 secs)

= 2100 secs

speed = distance/time

V = 5.508 × 10¹¹m / 2100 s

V = 2.62 × 10⁸ m/s

Answer:

no the only things that can travel at the speed of light are waves in the electromagnetic spectrum

Answer:

The largest  possible distance is [tex]x = 4.720 \ m[/tex]

Explanation:

From the question we are told that

    The distance of  separation is   [tex]d = 4.30 \ m[/tex]

      The  frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The  wavelength of the tone played by the speaker is  mathematically evaluated as

              [tex]\lambda = \frac{v}{f}[/tex]

substituting values

            [tex]\lambda = \frac{343}{103}[/tex]

            [tex]\lambda = 3.33 \ m[/tex]

Let the the position of the observer be O

Given that the line of sight between observer and speaker B is  perpendicular to the distance between A and B then

        The distance between A and the observer is  mathematically evaluated using Pythagoras theorem as follows

               [tex]L = \sqrt{d^2 + x^2}[/tex]

Where x is the distance between the observer and B

  For the observer to observe destructive interference

          [tex]L - x = \frac{\lambda}{2}[/tex]

So  

          [tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]

       [tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]

        [tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]

         [tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]

       [tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]

substituting values              

       [tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]

      [tex]x = 4.720 \ m[/tex]

Answer:

The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.

Explanation:

The riders will experience a centripetal force from the cylinder

[tex]F_{C}[/tex] = mrω^2    .... equ 1

where

m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of of the rider

For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as

[tex]F_{f}[/tex] = μR       ....equ 2

where

μ = coefficient of friction = 0.87

R is the normal force from the rider = mg

where

m is the rider's mass

g is the acceleration due to gravity = 9.81 m/s

substitute mg for R in equ 2, we'll have

[tex]F_{f}[/tex] = μmg     ....equ 3

Equating centripetal force of equ 1 and frictional force of equ 3, we'll get

mrω^2 = μmg

the mass of the rider cancels out, and we are left with

rω^2 = μg

ω^2 = μg/r

ω = [tex]\sqrt{\frac{ug}{r} }[/tex]

ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]

ω = 1.58 rad/second

The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s

The riders will experience a  centripetal force from the cylinder

[tex]F = mrw^2[/tex]

where  m is the mass of the rider

r is the inner radius of the cylinder = 3.4 m

ω is the angular speed of the rider

For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as

f = μN

where

μ = coefficient of friction = 0.87

N is the normal force = mg

f = μmg  

Equating centripetal force of and frictional force of we'll get

[tex]mrw^2 = umg[/tex]

[tex]rw^2 = ug[/tex]

[tex]w^2 = ug/r[/tex]

[tex]w= \sqrt{ug/r}[/tex]

[tex]w= \sqrt{0.87*9.8/3.4}[/tex]  

ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.

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Answer:

Objects act differently in a gravity field than in an accelerating reference frame.

Explanation:

The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.

The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.

Answer:

A - The orbital path of mercury around the sun has changed.

Explanation:

got right on edg.

Answer:

at the top of the 9 story building i think

Explanation:

When the ball starts to move, its kinetic energy increases and potential energy decreases. Thus the ball will experience its maximum potential energy at the top height before falling.

Potential energy of a massive body is the energy formed by virtue of its position and displacement. Potential energy is related to the mass, height and gravity as P = Mgh.

Where, g is gravity m is mass of the body and h is the height from the surface.  Potential energy is directly proportional to mass, gravity and height.

Thus, as the height from the surface increases, the body acquires its maximum potential energy. When the body starts moving its kinetic energy progresses and reaches to zero potential energy.

Therefore, at the sate where the ball is at the  top of the building it have maximum potential energy and then changes to zero.

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Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

Explanation:

Answer:

i

  [tex]v = 142.595 \ m/s[/tex]

ii

  [tex]f = 109.69 \ Hz[/tex]

iii1 )

  [tex]f_2 =219.4 Hz[/tex]

iii2)

   [tex]f_3 =329.1 Hz[/tex]

iii3)

    [tex]f_4 =438.8 Hz[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]l = 0.65 \ m[/tex]

     The tension on the string is  [tex]T = 61 \ N[/tex]

     The mass per unit length is  [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]

The speed of wave on the string is mathematically represented as

       [tex]v = \sqrt{\frac{T}{m} }[/tex]

substituting values

      [tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]

     [tex]v = 142.595 \ m/s[/tex]

generally the  string's  frequency is mathematically represented as

         [tex]f = \frac{nv}{2l}[/tex]

n = 1  given that the frequency we are to find is the fundamental frequency

So

      substituting values

       [tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]

       [tex]f = 109.69 \ Hz[/tex]

The  frequencies at which the string would vibrate include

1       [tex]f_2 = 2 * f[/tex]

Here [tex]f_2[/tex] is  know as the second harmonic and the value is  

      [tex]f_2 = 2 * 109.69[/tex]

      [tex]f_2 =219.4 Hz[/tex]

2

[tex]f_3 = 3 * f[/tex]

Here [tex]f_3[/tex] is  know as the third harmonic and the value is  

      [tex]f_3 = 3 * 109.69[/tex]

     [tex]f_3 =329.1 Hz[/tex]

3

     [tex]f_3 = 4 * f[/tex]

Here [tex]f_4[/tex] is  know as the fourth harmonic and the value is  

      [tex]f_3 = 4 * 109.69[/tex]

     [tex]f_4 =438.8 Hz[/tex]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

Answer:

Total heat transfer is positive

Total work transfer is positive

Explanation:

The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.

Heat transfer to a system is positive and that transferred from the system is negative.

Also, work done by a system is positive while the work done on the system is negative.

Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)

Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).

Answer:

The center of mass velocity is  [tex]v = 32.25 \ m/s[/tex]

Explanation:

From the question we are told that

          The mass of the cylinder is  [tex]m = 2.50 \ kg[/tex]

            The radius  is  [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]

             The length is  [tex]l = 2.7 \ cm = 0.027 \ m[/tex]

              The height of the plane is  h  = 79.60  m

               and the distance covered is  [tex]d = 4.64 \ m[/tex]

The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as

              [tex]v = \sqrt{\frac{4gh}{3} }[/tex]

substituting values  

               [tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]

              [tex]v = 32.25 \ m/s[/tex]

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

Answer:

b

Explanation:

because:/

Answer:

radius of the loop =  7.9 mm

number of turns N ≅ 399 turns

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = [tex]4\pi *10^{-7} T-m/A[/tex] = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = [tex]2\pi r[/tex]

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = [tex]7.9*10^{-4} m[/tex] = 0.0079 m = 7.9 mm

Answer:

the energy of the photons is greater than the work function of the zinc oxide.

                     h f> = Ф

Explanation:

In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.

                    K_max = h f - Ф

in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.

                     h f > = Ф

Answer:

k = 0.5 MN/m

Explanation:

Mass of the railcar, m = 5000 kg

Speed of the rail car, v = 1 m/s

The Kinetic energy(KE) of the railcar is given by the equation:

KE = 0.5 mv²

KE = 0.5 * 5000 * 1²

KE = 2500 J

The spring's compression, x = 0.1 m

The potential energy(PE) stored in the spring is given by the equation:

PE = 0.5kx²

PE = 0.5 * k * 0.1²

PE = 0.005k

According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring

KE = PE

2500 = 0.005k

k = 2500/0.005

k = 500000 N/m

k = 0.5 MN/m

Answer:

Explanation:

Given data

mass of  load m= 425 kg

height/distance h=64 m

acceleration a= 1.8 m/s^2

The work done can be calculated using the expression

work done= force* distance

but force= mass *acceleration

hence work done= 425*1.8*64= 48,960 J

work done= 48.96 kJ

Answer:

0.8 m

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing towards the center.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the centripetal direction:

∑F = ma

Nμ = m v²/r

Substitute and simplify:

mgμ = m v²/r

gμ = v²/r

Write v in terms of ω and solve for r:

gμ = ω²r

r = gμ/ω²

Plug in values:

r = (10 m/s²) (0.5) / (2.5 rad/s)²

r = 0.8 m

The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.

Given the following data:

To determine how far (radius) from the axis of rotation can the man stand without sliding:

We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.

[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.

But, Normal force, [tex]F_n = mg[/tex]  

Substituting the normal force into eqn. 1, we have:

[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.

Also, Linear speed, [tex]v = r\omega[/tex]

Substituting Linear speed into eqn. 2, we have:

[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]

Radius, r = 0.784 meters

Read more: https://brainly.com/question/13754413

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c