An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.

Answer:

5.79 in

Explanation:

We are given that

Diameter,d=0.30 in

Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]

Weight of hydrometer,W=0.042 lb

Specific gravity(SG)=1.10

Height of stem from the water surface=3.15 in

Density of water=[tex]62.4lb/ft^3[/tex]

In water

Volume  of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]

Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]

Change in volume=V-V'

[tex]V-V'=\pi r^2 l[/tex]

Substitute the values

[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]

By using

1 ft=12 in

[tex]\pi=3.14[/tex]

[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]

l=2.64 in

Total height=h+l=3.15+2.64= 5.79 in

Hence, the height of the stem protrude above the liquid surface=5.79 in

Answer:

Explanation:

As a result of impact of time widening, a clock moving as for an observer seems to run all the more gradually than a clock that is very still in the observer's casing.  

At the point when observed from earth, the pendulum on the spaceship takes more time to finish one oscillation.  

Hence, the clock related with that pendulum will run more slow (gives fewer oscillations as observed from the earth)  than the clock related with the pendulum on earth.

Ans => B fewer oscillations

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0  m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]

So, the plane slide a distance of 229.5 m.  

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

Substitute these values into equation (i) as follows;

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² +  [tex]\frac{1}{2}[/tex]m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) +  ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

Answer:

  m = 4

Explanation:

The expression that explains the constructive interference of a diffraction pattern is

         a sin θ = m λ

where a  is the width of the slit and λ the wavelength

         sin θ = m λ / a

The maximum value is for when the sine is 1, let's substitute

         1 = m λ/a  

         m = a /λ

let's reduce the magnitudes to the SI system

        a = 2.1 um = 2.1 10⁻⁶

        lam = 475 nm = 475 10⁻⁹ m

let's calculate

        m = 2.1  10⁻⁶ / 475 10⁻⁹

        m = 4.42

with m must be an integer the highest value is

         m = 4

Answer:

1.363×10^15 seconds

Explanation:

The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.

To calculate the time taken from Kepler's third Law :

T^2 = ( 4π^2/GMe ) r^3

Where Me is the mass of the earth

r is the average distance travel

G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2

π = 3.14

Me = mass of earth = 5.972×10^24kg

r =( r minimum + r maximum)/2 ......1

rmin = 6590km

rmax = 385000km

From equation 1

r = (6590+385000)/2

r = 391590/2

r = 195795km

From T^2 = ( 4π^2/GMe ) r^3

T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3

= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15

= 39.4384/ 3.983×10^14 ) × 7.5059×10^15

= (9.901×10^14) × 7.5059×10^15

T^2 = 7.4321× 10^30

T =√7.4321× 10^30

T = 2.726×10^15 seconds

The time for one way trip from Earth to the moon is :

∆T = T/2

= 2.726×10^15 /2

= 1.363×10^15 secs

Answer:

The final velocity of the two players is 0.69 m/s in the direction of the opposing player.

Explanation:

Since the players are moving in opposite directions, from the principle of conservation of linear momentum;

[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v

Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.

[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;

103 × 2.0 - 110 × 3.2 = (103 + 110)v

206 - 352 = 213 v

-146 = 213 v

v = [tex]\frac{-146}{213}[/tex]

v = -0.69 m/s

The final velocity of the two players is 0.69 m/s in the direction of the opposing player.

Answer:

8.167

Explanation:

Answer:

ML²/6

Explanation:

Pls see attached file

The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,

The length of the rod is L. It has a non-uniform distribution of mass given by:

dm/dx = Cx

where C has units kg/m²

dm = Cxdx

the total mass M of the rod can be calculated by integrating the above relation over the length:

[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]

Thus,

C = 2M/L²

Now, the moment of inertia of the small element dx of the rod is given by:

dI = dm.x²

dI = Cx.x²dx

[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]

I = ML²/2

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Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

Answer:

v = 8.5 m/s

Explanation:

In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,

Weight = Centripetal Force

but,

Weight = mg

Centripetal Force = mv²/r

Therefore,

mg = mv²/r

g = v²/r

v² = gr

v = √gr

where,

v = minimum speed required = ?

g = 9.8 m/s²

r = radius = 7.4 m

Therefore,

v = √(9.8 m/s²)(7.4 m)

v = 8.5 m/s

Minimum speed for a roller coaster while travelling upside down  so that the person will not fall out = 8.5 m/s

For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.

In the given condition the weight of the person must be balanced by the centrifugal force.

and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person

According to the Newton's Second Law of motion we can write force balance

[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]

Given Radius of loop = r = 7.4 m

Putting the value  of r = 7.4 m  in equation (1) we get

[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]

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Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.

Answer:

The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

Explanation:

The terminal velocity of the bacterium can be calculated using the following equation:

[tex] F = 6\pi*\eta*rv [/tex]    (1)

Where:

F: is drag force equal to the weight

η: is the viscosity = 1.002x10⁻³ kg/(m*s)

r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm

v: is the terminal velocity

Since that F = mg and by solving equation (1) for v we have:

[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]  

We can find the mass as follows:

[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]

Where:

ρ: is the density of the bacterium = 1.10x10³ kg/m³

V: is the volume of the spherical bacterium

[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]

Now, the terminal velocity of the bacterium is:

[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]

Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

I hope it helps you!

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

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

At constant speed, we will have;

[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

Answer:

8.7 m/s^2

82.15°

Explanation:

Given:-

- The initial speed of the car, vi = 25 m/s

- The radius of track, r = 129 m

- Car makes a circular " quarter turn "

- The constant tangential acceleration, at = 1.2 m/s^2

Solution:-

- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:

                          [tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]

- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:

                           at = r*α

                           α = ( 1.2 / 129 )

                           α = 0.00930 rad/s^2

- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).

- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:

                            [tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]

- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:

                            [tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]

- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:

                            [tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]

- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:

                             [tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]

- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:

                          [tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex] 

Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .

Answer:

1.6 m/s west

Explanation:

The recoil velocity of the launcher is 1.6 m/s west.

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of 8 m/s to the east.

Final momentum will be zero, so

m₁u₁ +m₂u₂ =0

Substitute the values for m₁ = 5kg, m₂ =1kg and u₂ =8 m/s, then the recoil velocity will be

5 x v +1x8 = 0

v = - 1.6 m/s

Thus, the recoil velocity of the launcher is  1.6 m/s (West)

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Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

Answer:

It always contains certain elements

Explanation:

Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.

Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.

The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

A mineral is a naturally occurring chemical compound, usually of a crystalline form.

Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

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Answer:

Explanation:

Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.

q = 1.6 x 10-19 C

v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s

the force acting on electron is

F= qvBsinΦ

F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)

F = 1.793x 10⁻¹⁸ N

The net force acting on electron is

F = e ( E+ ( vXB)

= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)

= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)

a= F/m

1.60 × 10¹² i =  ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹

9.11 i = - ( E- 6.7 k + 9.0 j)

E = -9.11i + 6.7k - 9.0j

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

Answer:

moves into a region of higher potential

Potential difference = 835   V

Ki = 835 eV

Explanation:

given data

initial speed = 400000 m/s

solution

when proton moves against a electric field  so that it will move into higher potential  region

and

we know Work done by electricfield  W is express as

W = KE of proton   K

so

q × V   =  0.5 × m × v²     ......................1

put here va lue

1.6 × [tex]10^{-19}[/tex] × V   =   0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²

Potential difference V = 1.336 × 10-16 / 1.6  × 10-19      

Potential difference = 835   V

and

KE of proton in eV is express as

Ki  =   V numerical

Ki = 835 eV

Answer:a

Explanation:

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]

Where

[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.

[tex]F(x)[/tex] - Force as a function of position, measured in newtons.

Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]

[tex]k = 250\,\frac{N}{m}[/tex]

Now, the work function is obtained:

[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]

[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]

[tex]W = 0.313\,J[/tex]

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:

[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]

[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]

By using trigonometrical identities, the integral is further simplified:

[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]

[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]

[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]

[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]

[tex]A = 4\pi[/tex]

The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex]       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]

The interaction time to avoid that the water balloon breaks is 0.029s

Answer:

C) It is either ferromagnetic or paramagnetic

Explanation:

The complete question is given below

We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?

A) It must be ferromagnetic.

B) It must be paramagnetic.

C) It is either ferromagnetic or paramagnetic.

D) It must be diamagnetic.

A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.

Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.

The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.

In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.

Answer:

The permittivity of rubber is  [tex]\epsilon = 8.703 *10^{-11}[/tex]

Explanation:

From the question we are told that

     The  magnitude of the point charge is  [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]

      The diameter of the rubber shell is  [tex]d = 32 \ cm = 0.32 \ m[/tex]

       The Electric field inside the rubber shell is  [tex]E = 2500 \ N/ C[/tex]

The radius of the rubber is  mathematically evaluated as

              [tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         [tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]

Where [tex]\epsilon[/tex] is the permittivity of rubber

    =>     [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]

   =>      [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]

substituting values

            [tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]

            [tex]\epsilon = 8.703 *10^{-11}[/tex]

Answer:

Increase of volume (F)  = 8.01%

Explanation:

Given:

Density of fish = 1,080 kg/m³

Density of water = 1,000 kg/m³

density of air = 1.28 kg/m³

Find:

Increase of volume (F)

Computation:

1,080 kg/m³  + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³  

1,080 + 1.28 F =1,000 F + 1,000

80 = 998.72 F

F = 0.0801 (Approx)

F = 8.01%  (Approx)

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

Answer:

Explanation:

The index of refraction of the liquid can be computed from ...

  [tex]n_i\sin{(\theta_t)}=n_t[/tex]

where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.

For the given incidence angles, the computed indices of refraction are ...

  A: n = 1.52sin(52.0°) = 1.198

  B: n = 1.52sin(44.3°) = 1.062

  C: n = 1.52sin(36.3°) = 0.900

Answer:

The ratio  is  [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              [tex]I = \frac{2}{3} * m r^2[/tex]

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       [tex]RE = \frac{1}{2}* I * w^2[/tex]

Where  [tex]w[/tex] is the angular velocity which is mathematically represented as

             [tex]w = \frac{v}{r}[/tex]

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

So

             [tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]

            [tex]RE = \frac{1}{3} * mv^2[/tex]

Generally the transnational  kinetic energy of this motion is  mathematically represented as

                [tex]TE = \frac{1}{2} mv^2[/tex]

So  

      [tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]

       [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]