A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
time =

Answer:

The speed of the electron will be 6x10^8m/s

Explanation:

See attached file

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:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

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 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

Explanation:

it transforms electrical energy into mechanical energy.

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:

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:

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:

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

Answer:

5. The amount of work is the same for both.

Explanation:

Work done is a measure of change in kinetic energy of each egg

For both egg , the initial speed and mass are same , so they have equal initial Kinetic energy

For both egg , the final speed is 0 and mass are same , so they have equal final Kinetic energy which is 0.

So work done is same for both eggs since they have same change in kinetic energy.

Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]

For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

[tex]I_{c} = \int {J} \, dA[/tex]

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]

[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]

For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]

(b) Ampere's Law to calculate magnetic field B is given by:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

(i) First, first find [tex]I_{c}[/tex] for r ≤ R:

[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]

[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]

[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]

[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]

Calculating B(r), using Ampere's Law:

[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]

[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0

B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0

B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0

(ii) For r ≥ R:

[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]

So, as calculated before:

[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]

[tex]I_{c} =[/tex] I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0

For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.

A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:

The Radius of the cross-section of the wire R

Current passing through the wire I

Current Density [tex]J = \alpha r[/tex]

Constant [tex]\alpha[/tex]

Distance of the point from the center [tex]r[/tex]

For part a)

Consider a circular strip between two concentric circles of radii r and r+dr.

Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]

[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]

Integration

[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]

For part b)

The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.

We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as

[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]

where ‘I’ is the threading current through the circle of radius ‘r’

[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]

(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where

[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]

[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)

[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]

Find out more about the density here:

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Answer:

a.    σ = 3.82*10^-18C/m^2

b.    d = 2.00m

Explanation:

a. In order to calculate the surface charge density of the sheet, you first calculate the acceleration of the electron on its motion.

You use the following formula:

[tex]v^2=v_o^2-2ad[/tex]                (1)

v: final speed of the electron = 0m/s

vo: initial speed of the electron = 390m/s

a: acceleration of the electron = ?

d: distance traveled by the electron = 2.00m

You solve the equation (1) for a, and replace the values of all parameters:

[tex]a=\frac{v_o^2-v^2}{2d}\\\\a=\frac{(390m/s)^2}{2(2.00m)}=3.8*10^4\frac{m}{s^2}[/tex]

Next, you calculate the electric field that exerts the electric force on the electron, by using the second Newton law, as follow:

[tex]F_e=qE=ma[/tex]               (2)

q: charge of the electron = 1.6*10^-19C

E: electric field of the sheet = ?

m: mass of the electron = 9.1*10^-31kg

You solve the equation (2) for E:

[tex]E=\frac{ma}{q}=\frac{(9.1*10^{-31}kg)(3.8*10^{4}m/s^2)}{1.6*10^{-19}C}\\\\E=2.16*10^{-7}\frac{N}{C}[/tex]

Next, you use the following formula to calculate the surface charge density, by using the value of its electric field:

[tex]E=\frac{\sigma}{2\epsilon_o}[/tex]          (3)

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

σ: surface charge density of the sheet

You solve for σ:

[tex]\sigma=2\epsilon_o E=2(8.85*10^{-12}C^2/Nm^2)(2.16*10^{-7}N/C)\\\\\sigma=3.82*10^{-18}\frac{C}{m^2}[/tex]

The surface charge density of the sheet id 3.82*10^-18C/m^2

b. To calculate the required distance for the electron reaches the sheet, you take into account that the electron acceleration is the same in all places near the sheet, then by the result of the previous point, you can conclude that the electron must be fired from a distance of 2.00m.

Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

2.16 × 10^-16N

Explanation:

The computation of the net force is shown below:

Data provided in the question

Electron mass = 9.11 × 10^-31kg

V_o = 0

V_f =  3.00 × 10^6 m/s

s = 1.90 cm i.e 1.9 × 10^-2

Based on the above information, the force is

As we know that

[tex]Force\ f = ma = \frac{mv^2}{2s}\\\\ = \frac{(9.11\times 10^{-31})(3\times 10^{6})^2}{2(1.9\times 10^{-2})}[/tex]

= 2.16 × 10^-16N

Hence, the last option is correct

Basically we applied the above formula to determine the net force

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Answer:

198.9 x 10^-16

Explanation:

E = hc/ wavelength

E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)

E = 198.9 x 10^-16

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:

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:

V = 2.95*10^-3 V

E = 0 N/C

Explanation:

In order to calculate the electric potential due to the two protons at the midpoint between them, you use the following:

[tex]V=V_1+V_2=k\frac{q}{r}+k\frac{q}{r}=2\frac{kq}{r}[/tex]       (1)

where you have taken into account each contribution to the total electric potential, produced by each proton.

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q: charge of the proton = 1.6*10^{-19}C

r: distance from the point (at the midway distance between the protons) to one proton = 973nm = 973nm/2 = 486.5nm = 486.5*10^-9m

You replace the values of the parameters in the equation (1):

[tex]V=2\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{486.5*10^{-9}m}\\\\V=2.95*10^{-3}V[/tex]

The electric potential is 2.95*10^-3V

The electric field generated at the midpoint in between the protons is zero, because the electric field generated by each proton has the same magnitude but opposite direction.

E = 0N/C

Answer:

q/6Eo

Explanation:

See attached file pls

Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

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]

The heat emitted from anything is carried in the form of infrared waves. (C)

Answer:

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

Answer:

The angular displacement is  [tex]\theta = 28.33 \ rad[/tex]

Explanation:

From the question we are told that

     The speed of the driver is  [tex]v =13 \ m/ s[/tex]

     The acceleration of the driver is  [tex]a = 1.02 \ m/s^2[/tex]

      The time taken is [tex]t = 0.70 \ s[/tex]

      The radius of the tire is  [tex]r = 33 cm = 0.33 \ m[/tex]

The distance covered by the car during this  acceleration can be  calculated using the equation of motion as follows

        [tex]s = v*t +\frac{1}{2} * a * t^2[/tex]

Now substituting values  

       [tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]

      [tex]s = 9.35 \ m[/tex]

Now the angular displacement of the car with respect to the tire movement can be  represented mathematically as

      [tex]\theta = \frac{s}{r}[/tex]

substituting values

      [tex]\theta = \frac{9.35}{0.33}[/tex]

      [tex]\theta = 28.33 \ rad[/tex]

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.

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:

The change in potential energy is  [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  [tex]E = 950 \ N/C[/tex]

      The  distance traveled by the electron is  [tex]x = 2.50 \ m[/tex]

Generally the force on this electron is  mathematically represented as

     [tex]F = qE[/tex]

Where F is the force and  q is the charge on the electron which is  a constant value of  [tex]q = 1.60*10^{-19} \ C[/tex]

    Thus  

      [tex]F = 950 * 1.60 **10^{-19}[/tex]

      [tex]F = 1.52 *10^{-16} \ N[/tex]

Generally the work energy theorem can be mathematically represented as

          [tex]W = \Delta KE[/tex]

Where W is the workdone on the electron by the  Electric field and  [tex]\Delta KE[/tex]  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        [tex]W = F* x *cos( \theta )[/tex]

Where  [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis  

        So

             [tex]\Delta KE = F * x cos (0)[/tex]

substituting values

         [tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]

          [tex]\Delta KE = 3.8*10^{-16} J[/tex]

Now From the law of energy conservation

       [tex]\Delta PE = - \Delta KE[/tex]

Where [tex]\Delta PE[/tex] is the change  in  potential energy  

Thus  

        [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]

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