A 30 W engine generates 3600 J of energy. How long did it run for?

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:

2P

Explanation:

See attached file

Answer:

R = 1138.9 Ω

Explanation:

Hello,

In this case, for the given power (P) and current (I), we can compute the resistance (R) via:

R = P / I²

Thus, we obtain:

R = 1.64x10⁵ W / (12.0 A)²

R = 1138.9 Ω

Best regards.

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Answer:

The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.

Explanation:

Answer:

Key points that can be found in the realist philosophical position​ are as follows:

Answer:

Order of 10^(-35) m.

Explanation:

The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.

Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.

This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm

(a) 7.392cm

(b) -15.32 cm/s

(c) -184cm/s²

(d) 0.4πs and 8.00cm

The general equation of a simple harmonic motion (SHM) is given by;

x(t) = A cos (wt + Φ)        --------------(i)

Where;

x(t) = position of the body at a given time t

A =  amplitude or maximum displacement during oscillation

w = angular velocity

t = time

Φ = phase constant.

Given from question:

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

(a) At time t = 0;

The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;

x(0) = 8.00 cos (5(0) + π / 8)

x(0) = 8.00 cos (π /8)

x(0) = 8.00 x 0.924

x(0) = 7.392 cm

Therefore, the position of the piston at time t = 0 is 7.392cm

(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;

v(t) = [tex]\frac{dx(t)}{dt}[/tex]

v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]

v(t) = 8 (-5 sin (5t + π / 8))

v(t) = -40sin(5t + π / 8)     --------------------(iii)

Now, substitute t=0 into the equation as follows;

v(0) = -40 sin(5(0) + π / 8)

v(0) = -40 sin(π / 8)

v(0) = -40 x 0.383

v(0) = -15.32 cm/s

Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s

(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]

a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]

a(t) = -200 cos (5t + π / 8)

Now, substitute t = 0 into the equation as follows;

a(0) = -200 cos (5(0) + π / 8)

a(0) = -200 cos (π / 8)

a(0) = -200 x 0.924

a(0) = -184.8 cm/s²

Therefore, the acceleration of the piston at time t = 0 is -184cm/s²

(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;

x(t) = A cos (wt + Φ)                   --------------(i)

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

From these equations it can be deduced that;

Amplitude, A = 8.00cm

Angular velocity, w = 5 rads/s

But;

w = [tex]\frac{2\pi }{T}[/tex]           [Where T = period of oscillation]

=> T = [tex]\frac{2\pi }{w}[/tex]

=> T = [tex]\frac{2\pi }{5}[/tex]

=> T = 0.4π s

Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm

Answer:

θ = 33.8

a = 3.42 m/s²

Explanation:

given data

mass m = 2 kg  

coefficient of static friction μs = 0.67

coefficient of kinetic friction μk = 0.25

solution

when block start slide

N = mg cosθ    .............1

fs = mg sinθ   ...............2

now we divide equation 2 by equation 1 we get

[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]

[tex]\frac{\mu s N }{N}[/tex]  = tanθ

put here value we get

tan θ = 0.67

θ = 33.8

and

when block will slide  then we apply newton 2nd law

mg sinθ - fk = ma    ...............3

here fk = μk N = μk mg cosθ

so from equation 3 we get

mg sinθ -  μk mg cosθ = ma

so a will be

a = (sinθ - μk cosθ)g

put here value and we get

a = (sin33.8 - 0.25 cos33.8) 9.8

a = 3.42 m/s²

Answer:

The moment of inertia of the merry-go round is 453.125 kg.m²

Explanation:

Given;'

angular velocity of the merry-go round, ω = 1.6 rad/s

rotational kinetic energy, K =  580 J

Rotational kinetic energy is given as;

K = ¹/₂Iω²

Where;

I is the moment of inertia of the merry-go round

[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]

Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²

Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]

Given the following data:

To calculate the moment of inertia of the small merry-go round:

Mathematically, the rotational kinetic energy of an object is giving by the formula:

[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]

Where:

Making moment of inertia (I) the subject of formula, we have:

[tex]I = \frac{2E_{rotational}}{w^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]

Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]

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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 answer is A

Explanation:

Its A because he created a telescope to be able to observe stars.

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  [tex]P_{avg} = F * v_{avg}[/tex]

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            [tex]P_{avg} = \frac{W }{\Delta t }[/tex]

Where W is  is the Workdone which is  mathematically represented as

         [tex]W = F * s[/tex]

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                [tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]

Now this  displacement can be represented mathematically as  

            [tex]s = v_{avg} * \Delta t[/tex]

Where [tex]v_{avg }[/tex] is the average  velocity and [tex]\Delta t[/tex] is the time  taken  

So  

            [tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]

=>         [tex]P_{avg} = F * v_{avg}[/tex]

Answer:

Pavg =  Fvavg

Explanation:

Since the P (power) done by the F (force) is:

P = Fs/t

and we are looking for the velocity, so then it would be:

P = Fv

with the average velocity the answer is:

Pavg =  Favg

If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v  is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).

Answer:

in the direction of the applied force

Explanation:

0.17T

When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e

[tex]F_{C}[/tex] = [tex]F_{M}[/tex]     --------------(i)

But;

[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]   [m = mass of the particle, r = radius of the path, v = velocity of the charge]

[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]

Substitute these into equation (i) as follows;

[tex]\frac{mv^2}{r}[/tex] = qvB

Make B subject of the formula;

B = [tex]\frac{mV}{qr}[/tex]            ---------------(ii)

Known constants

m = 1.67 x 10⁻²⁷kg

q = 1.6 x 10⁻¹⁹C

From the question;

v = 1.4 x 10⁷m/s

r = 0.85m

Substitute these values into equation(ii) as follows;

B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]

B = 0.17T

Therefore, the magnetic field strength is 0.17T

Answer:

The total work done by the person is given as = m g h

= 4.5kg x 9.8m/s²x1.2m

= 52.92J

This is the work done in moving the block in a vertical distance

However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.

Explanation:

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

The electromagnetic energy is 344.8 x10⁻⁸J/m³

We have to use the formula which says

Electromagnetic energy = energy intensity/ speed of light

We are given intensity as 1000 W/m²

Electromagnetic energy    = 1000/2.9 x 10⁸

                                             = 344.8 x 10⁻⁸J/m³

Therefore the electromagnetic energy is contained in each cubic meter will be  344.8 x 10⁻⁸J/m³

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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:

the ray of light should approach normal

Explanation:

When light passes through two means of different refractive index, it fulfills the equation

              n₁ sin  θ₁ = n₂ sin θ₂

where index 1 and 2 refer to each medium

In this problem, they tell us that light passes to a medium with a higher index, which is why

               n₁ <n₂

let's look for the angle in the second half

            sinθ₂ = n₁ /n₂  sin θ₁

            θ₂ = sin⁻¹ (n₁ /n₂  sin θ₁)

let's examine the angle argument the quantity n₁ /n₂ <1   therefore the argument decreases, therefore the sine and the angle decreases

Consequently the ray of light should approach normal

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

[tex]\omega=\omega_{0}+\alpha t[/tex]

Put the value in the equation

[tex]\omega=0.17+1.3\times1.7[/tex]

[tex]\omega=2.38(k)\ m/s[/tex]

We need to calculate the angular displacement

Using angular equation of motion

[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]

Put the value in the equation

[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]

[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]

[tex]\theta= 124.18^{\circ}[/tex]

We need to calculate the velocity at point A

Using equation of motion

[tex]v_{A}=v_{0}+\omega\times r[/tex]

Put the value into the formula

[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]

[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]

[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]

We need to calculate the acceleration at point A

Using equation of motion

[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]

Put the value in the equation

[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]

[tex]a_{A}=0.791j-0.851i[/tex]

[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]

Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]

(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]

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 magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

Answer:

Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.

Given Information:  

Radius of wire = r = 0.405 mm = 0.405×10⁻³ m

Length of wire = L = 15 m

Voltage = V = 12 V

Resistivity =  ρ = 100×10⁻⁸ Ωm

Required Information:  

Current = I = ?

Answer:  

Current = I = 0.412 A

Explanation:  

The current flowing through the wire can be found using Ohm's law that is

V = IR

I = V/R

Where V is the voltage across the wire and R is the resistance of the wire.

The resistance of the wire is given by

R = ρL/A

Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(0.405×10⁻³)²

A = 0.515×10⁻⁶ m²

So the resistivity of the wire is

R = ρL/A

R = (100×10⁻⁸×15)/0.515×10⁻⁶

R = 29.126 Ω

Finally, the current flowing through the wire is

I = V/R

I = 12/29.126

I = 0.412 A

Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.

Answer:

Majorly the electric field is reduced among other effect listed in the explanation

Explanation:

In capacitors the presence of di-electric materials

1. decreases the electric fields

2. increases the capacitance of the capacitors.

3. decreases the voltage hence limiting the flow of electric current.

 The di-electric material serves as an insulator between the metal plates of the capacitors

Answer:

Such limitations are given below.

Explanation:

Answer:

in CGS system G is denoted as gram

Answer:

The hydrostatic force on one end of the trough is 54994.464 N

Explanation:

Given;

liquid density, ρ = 810 kg/m³

side of the equilateral triangle, L = 8m

acceleration due to gravity, g =  9.8 m/s²

Hydrostatic force is given as;

H = ρgh

where;

h is the vertical height of the equilateral triangle

Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.

let the half side of the triangle = x

x = ⁸/₂ = 4m

The half section of the triangle forms a right angled triangle

h² = 8² - 4²

h² = 48

h = √48

h = 6.928m

F = ρgh

F = 810 x 9.8 x 6.928

F = 54994.464 N

Therefore, the hydrostatic force on one end of the trough is 54994.464 N

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]

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:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution