When a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder

Answer:

[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

[tex]B=\frac{\mu_oI}{2\pi r}[/tex]

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]

Then you have:

[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Answer:

About 3.06 seconds

Explanation:

[tex]v_f=v_o+at[/tex]

Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:

[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]

Hope this helps!

Answer:

x = 3.76 cm

y = 3.76 cm

Explanation:

This composite shape can be modeled as a square (7.2 cm × 7.2 cm) minus a quarter circle in the lower left corner (3.6 cm radius) and a right triangle in the upper right corner (3.6 cm × 3.6 cm).

The centroid of a square (or any rectangle) is at x = b/2 and y = h/2.

The centroid of a quarter circle is at x = y = 4r/(3π).

The centroid of a right triangle is at x = b/3 and y = h/3.

Build a table listing each shape, the coordinates of its centroid (x and y), and its area (A).  Use negative areas for the shapes that are being subtracted.

Next, multiply each coordinate by the area (Ax and Ay), sum the results (∑Ax and ∑Ay), then divide by the total area (∑Ax / ∑A and ∑Ay / ∑A).  The result will be the x and y coordinates of the center of mass.

See attached image.

Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?

Explanation:

Answer:

Analytical method.

Answer: A and D

a. Many objects move in arcs of circles or complete circles at times, and polar coordinates allowthe motion of such objects to be comprehended more easily. Some of Newton's laws in certaincases, such as calculation ofg from first principles, are much easier to calculate using polar coordinates.

b. "These coordinates are just used to confuse students.

c. ""Physics teachers are helping math teachers by getting students to practice theirtrigonometry.

d. ""In some cases, such as addition of forces, where a force magnitude is specified, it is simpler todescribe the forces in polar coordinates and be able to convert to the xy representation.

Explanation:

(A). Any force possessing a fixed magnitude and direction, it's always important to describe the for exactly as it is and also be able to convert it from one coordinate representation to another.

(D). Anything which involves radial motion(this is a motion along a radius) or motion along an arc of a circle or ellipse, this kind of motion is best explained and easily understood when in polar coordinates.

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

[tex]d_0=5m=500cm[/tex]

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm[/tex]

By the formula of magnification

[tex]\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm[/tex]

The height of the image formed is 18.89cm.

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.

1. T = 100 K

[tex]^{\circ}C=K-273[/tex]

Put T = 100 K

[tex]T=100-273=-173^{\circ} C[/tex]

A temperature of 100 K corresponds on a Celsius scale to (-173 °C)

2. T = 50 °C

[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]

So, At 50 °C, it corresponds to a Kelvin scale of 323 K.

Explanation:

Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod).  There are three forces on the pendulum:

Weight force mg at the center of the sphere,

Reaction force in the x direction at the pivot,

Reaction force in the y direction at the pivot.

Sum the torques about the pivot O.

∑τ = I d²θ/dt²

mg (L sin θ) = I d²θ/dt²

For small θ, sin θ ≈ θ.

mg L θ = I d²θ/dt²

Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.

If you wish, you can use parallel axis theorem to find the moment of inertia about O:

I = Icm + md²

I = ⅖ mr² + mL²

mg L θ = (⅖ mr² + mL²) d²θ/dt²

gL θ = (⅖ r² + L²) d²θ/dt²

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

[tex]t_A=0.55s[/tex]

[tex]t_B=0.45s[/tex]

(a) The distance from the kicker to each of the 2 spectators is given by:

[tex]d_A=v \times t_A[/tex]

where,

v= speed of sound

[tex]t_A[/tex]=time taken for the sound waves to reach the ears

[tex]d_A=343\times 0.55=188.65[/tex]m

(b)[tex]d_B=v \times t_B[/tex]

where,

v= speed of sound

[tex]t_B[/tex]=time taken for the sound waves to reach the ears

[tex]d_B=343\times 0.45=154.35m[/tex]

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]

Answer:

6 A

Explanation:

Parallel connected resistors needs to be calculated as one single resistor. To do that: [tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]=[tex]\frac{3}{15}[/tex]=[tex]R^{-1}[/tex]

[tex]\frac{3}{15} ^{-1}[/tex]= 5 Ω (total resistance)

U = R* I

[tex]\frac{U}{R}[/tex]=I

[tex]\frac{30}{5}[/tex]=6 A

Answer:

(a) 0.0885 Hz

(b) 0.21 m/s

Explanation:

(a) Frequency: This can  be defined as the number of cycle completed in one seconds.

From the question,

Note: 2 crest = one cycle,

If four crest = 22.6 s,

Then two crest = (22.6/2) s

= 11.3 s.

T = 11.3 s

But,

F = 1/T

F = 1/11.3

F = 0.0885 Hz.

(b)

Using,

V = λF...................... Equation 1

Where V = speed of wave, F = Frequency of wave, λ = wave length.

Given: F = 0.0885 Hz, λ = 2.37 m.

Substitute these values into equation 1

V = 2.37(0.0885)

V = 0.21 m/s.

Answer:

A) The two potatoes cover the same horizontal distance from the launcher.

B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.

C) Marvin's potato hits the ground with a greater speed than Mark's potato

Explanation:

A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched = 30°, 60°

g = acceleration due to gravity on Mars

Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.

Sin (2×60°) = sin 120°

Sin (2×30°) = sin 60°

Sin 120° = Sin 60°

Hence, the two potatoes cover the same horizontal distance from the launcher.

B) Time spent in the air for a projectile is given as

T = (2u sin θ)/g

Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.

Sin 60° = 0.866

Sin 30° = 0.50

Sin 60° > Sin 30°

Hence, Mark's potato spends more time in the air than Marvin's potato.

C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ

u cos 60° < u cos 30°

And the initial vertical velocity is u sin θ

Final vertical velocity

= (initial vertical velocity) - gt

g = acceleration due to gravity on Mars

T = time of flight

For Mark,

initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°

And Mark's potato's time of flight is greater as established in (B) above.

But for Marvin

initial vertical velocity = u sin 30°, less than Mark's u sin 60°

And Marvin's potato's time of flight is lesser as established in (B) above

So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.

Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.

Hope this Helps!!!

Answer:

e.) power you supply to the crate

Explanation:

According to given data, we have:

F = Force exerted on the crate

M = Mass of the crate

v = Speed of motion of the crate

d = Distance traveled by the crate across the floor

t = Time interval passed

Now, we try to analyze the given quantity:

=> F d/t

=> (Force)(Displacement)/(Time)

but, (Force)(Displacement) = Work Done

Therefore,

=> Work Done/Time

but, Work Done/Time = Power

Therefore,

=> Power

Hence, the quantity F d/t is:

e.) power you supply to the crate

Answer:

The curve should be banked at an angle of 13 degrees.

Explanation:

We have,

Radius of a highway curve is 274 m

Speed of car on this curve is 25 m/s

Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.

[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]

g = 10 m/s²

Plugging all the values in above formula,

[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]

So, the curve should be banked at an angle of 13 degrees.

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

Answer:

2500 N/m²

Explanation:

Pressure: This can be defined as the force acting normally on a surface per unit area.

The expression for pressure is give as

P = F/A...................... Equation 1

Where P = pressure (N/m²), F = force (N), A = Contact area (m²)

Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.

Substitute into equation  1

P = 500/0.2

P = 2500 N/m²

Hence the pressure exerted to the surface is 2500 N/m²

Answer:

Explanation:

Ex(z,t) = Eocos(kz - ω t + φ)

k = 2π/λ  , ω = 2π f

φ = +30° , E₀ = 10³ V .

z/λ = 0.25 , ft = 0.125

Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)

Putting the values given above

Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )

= 1000cos (90⁰ - 45+30)

= 1000 cos 75

=258.8  V .

Answer:

It will lose electrical potential energy.

Explanation:

A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.

Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.

Answer:

a.  4 V

b. 0.697 A

Explanation:

Magnetic field strength B =  0.732 T

length of rod l = 0.362 m

velocity of rod v = 15.1 m/s

a.  EMF can be calculated as

E = Blv = 0.732 x 0.362 x 15.1 = 4 V

b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω

current I = V/R = 4/5.74 = 0.697 A

the current flows in a clockwise direction

Answer:

The answer is D)Theory

Explanation:

This is due because a theory is a scientific term and is a testable model that scientists seek to explain a phenomenon. You can also find out the answer by the process of elimination it can't be data because that would be something they already know and something they use to prove not explain. It can't be law because it isn't testable but can be used to explain. So that leaves you with two answers hypothesis and theory which are very similar but it isn't hypothesis because it isn't used to explain it to help the scientists come up with a theory and accumulate what might happen.

Answer:

The work done by the force is  5.76 J

Explanation:

Given;

mass of canister , m = 3.2 kg

magnitude of force, f = 6.7 N

initial velocity of the canister on x-axis,  [tex]v_i[/tex]= 3.3i m/s

final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s

The work done on the canister = change in the kinetic energy of the canister

[tex]W = K.E_f - K.E_i[/tex]

where;

K.Ei is the initial kinetic energy

K.Ef is the final kinetic energy

The initial kinetic energy:

[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]

The final kinetic energy:

[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]

W = 11.04 - 5.28

W = 5.76 J

Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

[tex]y=\frac{1}{2}at^2[/tex]

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

[tex]y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m[/tex]

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s[/tex]

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}[/tex]

Answer:

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.

[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]

Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:

[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]

[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]

The angular accelaration measured in radians per square second is:

[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]

[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

[tex]\tau = I \cdot \alpha[/tex]

Where:

[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

Where:

[tex]m[/tex] - Mass of the CD, measured in kilograms.

[tex]r[/tex] - Radius of the CD, measured in meters.

If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:

[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]

[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]

Now, the net torque exerted on CD is:

[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]

[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Answer:

  B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

Answer:

a = 5.34 m/s²

T = 37.86 N

Explanation:

This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:

a = g(m₂ - m₁)/(m₂ + m₁)

where,

a = acceleration of both masses = ?

g = 9.8 m/s²

m₂ = heavier mass = 8.5 kg

m₁ = lighter mass = 2.5 kg

Therefore,

a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)

a = (9.8 m/s²)(6 kg)/(11 kg)

a = 5.34 m/s²

The formula for tension in cable is derived to be:

T = 2m₁m₂g/(m₁ + m₂)

T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)

T = 37.86 N

Answer:

The distance from the cornea vertex to the retina is 2.36 cm

Explanation:

The question is incomplete.

The complete question is as follows;

A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.

If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?

Solution.

We use image-object reaction to calculate the distance from the cornea vertex to the retina.

Mathematically;

n1/s + n2/s’ = n2-n1/R

From the question, we identify the following;

n1 ; Refractive index of air = 1

n2 ; Refractive index of lens = 1.4

S ; Object Distance = 36 cm

S’ = ?

R ; Radius of curvature of the cornea = 0.65

Substituting these values into the equation above;

1/36 + 1.4/S’ = (1.4-1)/0.65

{S’+ 36(1.4)}/36S’ = 0.4/0.65

{S’ + 50.4}/36S’ = 0.62

S’ + 50.4 = 22.32S’

50.4 = 22.32S’ -S’

21.32S’ = 50.4

S’ = 50.4/21.32

S’ = 2.36 cm