An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at the top 16.0°C, what is the volume of the bubble just before it reaches the surface?

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Where, [tex]\mu[/tex] = mass per unit length

T = tension

Put the value into the formula

[tex]v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}[/tex]

[tex]v =519.61\ m/s[/tex]

Hence, The speed of transverse waves in this string is 519.61 m/s.

Answer:

[tex] v_{max}=14.2\frac{m}{s} [/tex]

Explanation:

Hi!

If the crate is not sliding, its trajectory is the arc with 36.1 m radius. Then the crate  has a centripetal acceleration:

[tex]a_c= \frac{v^2}{r} \\r = radius\\v = tangential \; velocity[/tex]

The centripetal force acting on the crate is the static friction force between crate and truck. The maximum value of this force is:

[tex]F_{max} = \mu N\\\mu = 0.570=static\;friction \;coefficient\\N =normal\; force\\[/tex]

The normal force has a magnitude equal to the weight of the crate:

[tex]N=mg[/tex]

Then the condition for not sliding is:

[tex]F_{centripetal} = M\frac{v^2}{r}<\mu N=\mu Mg\\ v^2<r \mu g = 36.1\;m*0.570*9.8\frac{m}{s^2}= 201.65 \frac{m^2}{s^2}\\ v<14.2\frac{m}{s}[/tex]

Answer:

The life time of the particle is [tex]2.491\times 10^{- 25} s[/tex]

Solution:

As per the question:

Average rest energy of [tex]Z^{0}boson = 91.19 GeV[/tex]

Uncertainty in rest energy, [tex]\Delta E_{r} = 2.5 GeV = 2.5\times 10^{9}\times 1.6\times 140^{- 19} J = 4\times 10^{- 10} J[/tex]

Now,

From the Heisenberg's Uncertainty Principle, we can write:

[tex]\Delta E_{r}\times \Delta T \geq \frac{h}{2\pi}[/tex]

where

T = Life time  of the particle

[tex]\Delta T \geq \frac{h}{2\pi\Delta E_{r}}[/tex]

[tex]\Delta T \geq \frac{6.262\times 10^{- 34}}{2\pi\times 4\times 10^{- 10}}[/tex]

[tex]\Delta T \simeq 2.491\times 10^{- 25} s[/tex]

The lifetime of the Z0 boson is approximately 8.95 x 10^-17 seconds.

The Z0 boson is a particle that mediates the weak nuclear force. Its average rest energy is 91.19 GeV and it has an intrinsic width of 2.5 GeV. The lifetime of a particle can be calculated using the Heisenberg uncertainty principle, which relates the energy uncertainty to the time uncertainty. The relationship is given by the equation ΔE × Δt ≥ ℏ/2, where ℏ is the reduced Planck constant. By rearranging the equation and substituting the values, we can calculate the lifetime of the Z0 boson to be approximately 8.95 x 10^-17 seconds.

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Answer:

The angle is [tex]\theta\approx 14.61[/tex] degrees.

Explanation:

Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:

[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]

If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.

[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].

If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].

[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]

The angle is thus

[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.

Answer:

Height, h = 300.27 meters

Explanation:

Given that,

Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]

We need to find the height of the manometer required. The pressure at a height is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]

h is the height of the manometer required.

[tex]h=\dfrac{P}{\rho g}[/tex]

[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]

h = 300.27 meters

So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.

The ratio of the orbital periods (Ta/Tb) of two satellites, where the orbital speed of satellite A is twice that of satellite B, can be found using Kepler's third law, which relates the orbital period squared to the radius of the orbit cubed. By understanding the relationship between orbital speed and radius, the ratio of the orbital periods can be calculated.

When comparing two satellites, A and B, with orbital speeds such that the orbital speed of satellite A is twice that of satellite B, we are tasked with finding the ratio of their orbital periods (Ta/Tb). This problem is grounded in the principles of classical mechanics and specifically relates to Kepler's laws of planetary motion.

According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the radius of the orbit (r). Mathematically, this is expressed as T² ≈ r³ for a satellite orbiting a much larger body, such as the Earth. Since the gravitational attraction provides the necessary centripetal force for the satellite's circular motion, the gravitational force is also centrally involved in this relationship.

To compare the periods of two satellites, we use this proportionality. If the orbital speed of satellite A is twice that of satellite B, the radius of the orbit is also related to the speed. Specifically, speed is directly related to the square root of the radius of the orbit based on centripetal force considerations. Since the speed of satellite A is twice that of satellite B (VA = 2*VB), it follows that the radius of A's orbit would be four times that of B's orbity (rA = 4*rB). Applying Kepler's law then allows us to find the period ratio as (Ta/Tb)² = (rA/rB)³, and after substituting the relation for the radii, we can solve for (Ta/Tb).

Answer:

Explanation:

Capacity of a parallel plate capacitor  C = ε₀ A/ d

ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.

C =(  8.85 X10⁻¹² X  27 X 10⁻⁴ ) / 3 X 10⁻³

= 79.65 X 10⁻¹³ F.

potential diff between plate = Charge / capacity

= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³

= 601 V

Electric field = V/d

= 601 / 3 x 10⁻³

= 2 x 10⁵ N/C

Force on proton

= charge x electric field

1.6 x 10⁻¹⁹ x 2 x 10⁵

= 3.2 x 10⁻¹⁴

Acceleration a = force / mass

= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷

= 1.9 x 10¹³ m s⁻²

Distance travelled by proton = 3 x 10⁻³

3 x 10⁻³ = 1/2 a t²

t = [tex]\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }[/tex]

t = 1.77 x 10⁻⁸ s

Answer:

In order to use the parallelogram method, we have to select first two vectors.  We move the vectors until their initial points coincide. Then we draw lines to form a complete parallelogram, as is shown in the figure annexed. The diagonal from the initial point to the opposite vertex of the parallelogram is the resultant.

We use the last vector resultant with a third vector, and we use again the parallelogram method to add them. We use the new resultant and we add it with a 4th vector. We repeat this task until all the vectors are used.

Answer:

93750 ft/s²

Explanation:

t = Time taken

u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)

v = Final velocity = 0

s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]

a = Acceleration

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]

The magnitude of acceleration is 93750 ft/s²

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

[tex]v^2-u^2=2ah[/tex]

here Final Velocity v=0

a=acceleration due to gravity

[tex]0-u^2=2\left ( -g\right )h[/tex]

[tex]u=\sqrt{2gh}[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]

[tex]u=\sqrt{24.328}[/tex]

u=4.93 m/s

Answer:

a)[tex]E=2.88*10^{11}N/C[/tex]

b)[tex]E=1.44*10^{11}N/C[/tex]

c)[tex]F=4.61*10^{-8}N[/tex]

Explanation:

We use the definition of a electric field produced by a point charge:

[tex]E=k*q/r^2[/tex]

a)Electric Field  due to the alpha particle:

[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C[/tex]

b)Electric Field  due to electron:

[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C[/tex]

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N[/tex]

Answer:26.96 m,2.34 s

Explanation:

Given

Wrench hit the ground with a speed of 23 m/s

Applying equation of motion

[tex]v^2-u^2=2as[/tex]

Here u=0 because it is dropped from a height of S m

[tex]23^2-0=2\times 9.81\times s[/tex]

[tex]s=\frac{529}{2\times 9.81}=26.96 m[/tex]

Time required by wrench to hit the ground

v=u+at

[tex]23=9.81\times t[/tex]

[tex]t=\frac{23}{9.81}=2.34 s[/tex]

Answer:

astronauts age is 32 years

correct option is e 32 years

Explanation:

given data

travels = 20 light year

stay = 2 year

return = 52 years

to find out

astronauts aged

solution

we know here they stay 2 year so time taken in traveling is

time in traveling = ( 52 -2 )  = 50 year

so it mean 25 year in going and 25 years in return

and distance is given 20 light year

so speed will be

speed = distance / time

speed = 20 / 25 = 0.8 light year

so time is

time = [tex]\frac{t}{\sqrt{1-v^2} }[/tex]

time =  [tex]\frac{25}{\sqrt{1-0.8^2} }[/tex]

time = 15 year

so age is 15 + 2 + 15

so astronauts age is 32 years

so correct option is e 32 years

Answer:

 507599.78 J

Explanation:

Charge input = current x time

=15 x 53 x 60

= 47000 coulomb

increase in voltage

= 12.6 - 9 = 3.6

capacity of the battery C = Charge input / increase in voltage

= 47000 / 3.6 = 13055.55

energy of the capacitor = 1/2 CV²

Initial energy of car battery =  .5 x 13055.55 x 9 x 9

= 528749.77 J

Final energy of car battery

= .5 x 13055.55 x 12.6 x 12.6

= 1036349.55

Increase in energy = 507599.78 J

Final answer:

The total energy delivered to the car battery during its charging process from 9 V to 12.6 V over 53 minutes with a charging current of 15 A is 515,160 Joules.

Explanation:

To calculate the energy delivered to the car battery while it is being charged, we need to consider the change in voltage, the charge current, and the time it was charged. Since the voltage changed linearly from 9 V to 12.6 V over 53 minutes with a constant charging current of 15 A, we first convert the charging time to seconds (53 min  imes 60 s/min = 3180 s) and then use the formula Energy (E) = Power (P)  imes Time (t), where Power (P) is the product of voltage (V) and current (I).

Assuming a linear voltage increase, we take the average voltage (9 V + 12.6 V)/2 during the charge time. The average voltage is then 10.8 V. The energy delivered is given by:

E = P  imes t = V  imes I  imes t = 10.8 V  imes 15 A  imes 3180 s

E = 515160 Joules

Therefore, the energy delivered to the car battery during the charging process is 515,160 Joules.

Final answer:

To find the electric field strength between the disks, we can use Coulomb's law and the formula for electric field. The launch speed of the proton can be found using the conservation of energy.

Explanation:

To find the electric field strength between the disks, we can use the formula:

Electric Field = Force / Charge

The force between the disks is given by Coulomb's law:

Force = (k * q1 * q2) / r^2

Where k is the electrostatic constant (9 * 10^9 Nm^2/C^2), q1 and q2 are the charges on the disks, and r is the distance between them.

Substituting the given values, we have:

Force = (9 * 10^9 Nm^2/C^2) * (16 * 10^-9 C)^2 / (2.3 * 10^-2 m)^2

Calculating this expression, we get the force between the disks. Then, we can divide this force by the charge on either disk to find the electric field strength.

For part B, we can use the conservation of energy to find the launch speed of the proton. The potential energy difference between the disks can be calculated as:

Potential Energy = charge * voltage

Given the charge of the proton and the distance between the disks, we can find the voltage. Since the proton starts from rest, all of its initial potential energy will be converted into kinetic energy:

Kinetic Energy = (1/2) * mass * velocity^2

Where the mass of the proton is known.

Solving for velocity, we can find the launch speed of the proton.

The electric field measurements indicate a point charge located to the right of both measurement positions on the x-axis, being more than 5.00 m away. Using the electric field values (3.000 N/C at 2.00 m and 0.750 N/C at 5.00 m), and applying the equation E = kQ/r^2, we can solve for both the point charge's magnitude and location.

The student's question revolves around the electric field produced by a point charge and involves finding both the location of the point charge and its magnitude. To solve this, we need to apply Coulomb's Law and the principle that the intensity of the electric field (E) from a point charge can be described by the equation E = kQ/r^2, where k is Coulomb's constant (8.988  imes 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge to the point where the electric field is measured.

Given the electric field measurements at two different points both pointing to the left, we can deduce that the point charge is located to the right of both points on the x-axis. The point charge must thus be more than 5.00 m to the right of the origin.

By using the measured electric field values and the distances to set up two equations, we can determine the charge's value:

E1 = kQ/r1^2 (at 2.00 m)
E2 = kQ/r2^2 (at 5.00 m)

Where E1 = 3.000 N/C, E2 = 0.750 N/C, r1 = 2.00 m, and r2 = 5.00 m. By solving these equations simultaneously, we can find the value of Q and the exact location along the x-axis.

Answer:

The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].

Explanation:

Given that,

Distance = 1.50 m

Mass of one student = 70.0 kg

Mass of other student = 52.0 kg

We need to calculate the gravitational force

Using formula of gravitational force

[tex]F=\dfrac{Gm_{1}m_{2}}{r^2}[/tex]

Where, m₁ = mass of one student

m₂ = mass of other studen

r = distance between them

Put the value into the formula

[tex]F=\dfrac{6.67\times10^{-11}\times70.0\times52.0}{1.50^2}[/tex]

[tex]F=1.079\times10^{-7}\ N[/tex]

Hence, The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].

To calculate the gravitational force between the two students, we use Newton's law of universal gravitation, substituting the given values for mass and distance into the formula. The result suggests that the gravitational force would be incredibly small, aligning with our daily experiences.

The subject of this question is Physics, specifically gravitational force. From Newton's law of universal gravitation, we know that the gravitational force between two masses is given by the equation F = G(M₁M₂)/r², where F is the gravitational force, G is the gravitational constant, M₁ and M₂ are the two masses, and r is the distance between them.

Given that one student has a mass of 70 kg (M₁), the other a mass of 52 kg (M₂) and the distance between them is 1.5 m (r), we can substitute these values into the formula. Using a gravitational constant (G) of approximately 6.67 × 10-¹¹ Nm²/kg², the gravitational force (F) becomes:

F = (6.67 × 10-¹¹ Nm²/kg²)(70 kg)(52 kg)/(1.50 m)²

Note, though, that the gravitational force between two people sitting 1.50 m apart would be incredibly small due to the immense smallness of the gravitational constant. This is inline with our daily experiences where we don't feel any noticeable gravitational pull from an ordinary object.

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Answer:

[tex]\lambda=1.23[/tex]Å

Explanation:

Bragg's Law refers to the simple equation:

[tex]n\lambda = 2d sin(\theta)[/tex]

In this case:

n=1

θ = 12.062°

d=2.9344 Å

[tex]\lambda = 2*2.9344sin(12.062)=1.23[/tex]Å

Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, [tex]g=-9.8\ m/s^2[/tex]

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]v^2=2as+u^2[/tex]

[tex]v^2=2(-9.8)\times 1+(10)^2[/tex]

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

Answer:

8.96 m/s, upward direction

Explanation:

Given that, the initial velocity of the ball is,

[tex]u=10m/s[/tex]

And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,

[tex]a=9.8\frac{m}{s^{2} }[/tex]

And according to question vertical displacement is,

[tex]s=1m[/tex]

Now suppose v be the final velocity of the ball.

Applying third equation of motion,

[tex]v^{2}=u^{2}+2as[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement.

Substitute all the variables.

[tex]v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}[/tex]

Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.

Answer:

(C) a circle starting at time t=0 on the positive x axis

Explanation:

particle's position is

r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^

this is a parametric equation of a circle, because the axis at x and y are the same = R.

for t=0:

r=Ri^

so: circle starting at time t=0 on the positive x axis

On the other hand:

[tex]v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j][/tex]

The value of the magnitude of the acceleration is:

[tex]a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}[/tex]

we can recognise that this represent the centripetal acceleration.

Answer:

[tex]\theta=28.07^{\circ}[/tex]

Explanation:

Speed of van, v = 28 m/s

Radius of unbanked curve, r = 150 m

Let [tex]\theta[/tex] is the angle with the vertical. In case of banking of road,

[tex]T\ cos\theta=mg[/tex].............(1)

And

[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]..........(2)

From equation (1) and (2) :

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

[tex]tan\theta=\dfrac{(28)^2}{150\times 9.8}[/tex]

[tex]\theta=28.07^{\circ}[/tex]

So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.

Answer:

(a): 37.94 m.

(b): [tex]323.16^\circ.[/tex]

(c): 126.957 m.

(d): [tex]0.93^\circ.[/tex]

(e): 49.92 m.

(f): [tex]130.08^\circ.[/tex]

Explanation:

Given:

Any vector [tex]\vec A[/tex], making an angle [tex]\theta[/tex] with respect to the positive x-axis, can be written in terms of its x and y components as follows:

[tex]\vec A = A\cos\theta\ \hat i+A\sin\theta \ \hat j.[/tex]

where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along the x and y axes respectively.

Therefore, the given vectors can be written as

[tex]\vec a = 50\cos30^\circ \ \hat i+50\sin 30^\circ\ \hat j = 43.30\ \hat i +25\ \hat j\\\vec b = 50\cos195^\circ \ \hat i+50\sin 195^\circ\ \hat j = -48.29\ \hat i +-12.41\ \hat j\\\vec c = 50\cos 315^\circ \ \hat i+50\sin 315^\circ\ \hat j = 35.35\ \hat i +-35.35\ \hat j\\[/tex]

(a):

[tex]\vec a +\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29+35.35)\hat i+(25-12.41-35.35)\hat j\\=30.36\hat i-22.75\hat j.\\\\\text{Magnitude }=\sqrt{30.36^2+(-22.75)^2}=37.94\ m.[/tex]

(b):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }(\vec a + \vec b +\vec c)}{\text{y component of }(\vec a + \vec b +\vec c)}=\dfrac{-22.75}{30.36}=-0.749\\\Rightarrow \theta = \tan^{-1}(-0.749)=-36.84^\circ.[/tex]

The negative sign indicates that the sum of the vectors is [tex]36.84^\circ.[/tex] below the positive x axis.

Therefore, direction of this vector sum counterclockwise with respect to positive x-axis = [tex]360^\circ-36.84^\circ=323.16^\circ.[/tex]

(c):

[tex]\vec a -\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)-(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30+48.29+35.35)\hat i+(25+12.41-35.35)\hat j\\=126.94\hat i+2.06\hat j.\\\\\text{Magnitude }=\sqrt{126.94^2+2.06^2}=126.957\ m.[/tex]

(d):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }(\vec a - \vec b +\vec c)}{\text{y component of }(\vec a - \vec b +\vec c)}=\dfrac{2.06}{126.94}=0.01623\\\Rightarrow \theta = \tan^{-1}(0.01623)=0.93^\circ.[/tex]

(e):

[tex](\vec a + \vec b)-(\vec c + \vec d)=0\\(\vec a + \vec b)=(\vec c + \vec d)\\\vec d = \vec a + \vec b -\vec c.[/tex]

[tex]\vec d = \vec a +\vec b - \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)-(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29-35.35)\hat i+(25-12.41+35.35)\hat j\\=-40.34\hat i+47.94\hat j.\\\\\text{Magnitude }=\sqrt{(-40.34)^2+47.94^2}=62.65\ m.[/tex]

(f):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }\vec d}{\text{y component of }\vec d}=\dfrac{47.94}{-40.34}=-1.188\\\Rightarrow \theta = \tan^{-1}(-1.188)=-49.92^\circ.[/tex]

The x component of this vector is negative and y component is positive therefore the vector lie in second quadrant, which means, the direction of this vector, counterclockwise with respect to positive x axis = [tex]180^\cir.

c-49.92^\circ=130.08^\circ.[/tex]

Answer:

[tex]\Delta s\ =\ 21.33\ J/K[/tex]

Explanation:

Given,

Let T be the final temperature of the mixture,

Therefore From the law of mixing, heat loss by the water is equal to the heat gained by the ice,

[tex]m_iL_f\ +\ m_is_w(T_f\ -\ 0)\ =\ m_ws_w(T_w\ -\ T_f)\\\Rightarrow 333000\times 0.0138\ +\ 0.0138\times 4190T_f\ =\ 0.134\times 4190\times(70.7\ -\ T_f)\\\Rightarrow 4595.4\ +\ 57.96T_f\ =\ 39789.96\ -\ 562.8T_f\\\Rightarrow 620.76T_f\ =\ 35194.56\\\Rightarrow T_f\ =\ \dfrac{35194.56}{620.76}\\\Rightarrow T_f\ =\ 56.69^o\ C[/tex]

Now, We know that,

Change in the entropy,

[tex]\Delta s\ =\ s_f\ -\ s_i\ =\ \dfrac{Q}{T}\\\Rightarrow \displaystyle\int_{s_i}^{s_f} ds\ =\ \displaystyle\int_{T_i}^{T_f}\dfrac{msdT}{T}\\\Rightarrow \Delta s =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

Now, change in entropy for the ice at 0^o\ C to convert into 0^o\ C water.

[tex]\Delta s_1\ =\ \dfrac{Q}{T}\\\Rightarrow \Delta s_1\ =\ \dfrac{m_iL_f}{T}\ =\ \dfrac{0.0138\times 333000}{273.15}\ =\ 16.82\ J/K.[/tex]

Change in entropy of the water converted from ice  from [tex]273.15\ K[/tex] to water 330.11 K water.

From the equation (1),

[tex]\therefore \Delta s_2\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_2\ =\ 0.0138\times 4190\times \ln \left (\dfrac{273.15}{330.11}\ \right )\\\Rightarrow \Delta s_2\ =\ 6.88\ J/K[/tex]

Change in entropy of the original water from the temperature 342.85 K to 330.11 K

From the equation (1),

[tex]\therefore \Delta s_3\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_3\ =\ 0.134\times 4190\times \ln \left (\dfrac{330.11}{343.85} \right )\\\Rightarrow \Delta s_3\ =\ -2.36\ J/K[/tex]

Total entropy change = [tex]\Delta s\ =\ \Delta s_1\ +\ \Delta s_2\ +\ \Delta s_3\\\Rightarrow \Delta s\ =\ (16.82\ +\ 6.88\ -\ 2.36)\ J/K\\\Rightarrow \Delta s\ =\ 21.33\ J/K.[/tex]

Hence, the change in entropy of the system form then untill the system reaches the final temperature is 21.33 J/K

Answer:

Current through each resistor is 2 A.

Explanation:

Given that,

Voltage of a battery, V = 5 volts

Resistance 1, R = 1.25 ohms

Resistance 2,R' = 1.25 ohms

Both resistors are connected in series. The equivalent resistance is given by :

R" = R + R'

R" = 1.25 + 1.25

R" = 2.5 ohms

The current flowing throughout all resistors is same in series combination of resistors. Current can be calculated using Ohm's law as :

[tex]I=\dfrac{V}{R"}[/tex]

[tex]I=\dfrac{5\ V}{2.5}[/tex]

I = 2 A

So, the current through each resistor is 2 A. Hence, this is the required solution.

Answer:

75 °F

Explanation:

Air has a specific heat at constant pressure of:

Cpa = 0.24 BTU/(lbm*F)

The specific heat of water is:

Cpw = 1 BTU/(lbm*F)

The first law of thermodynamics:

Q = L + ΔU

The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.

Then:

Q = 0.

We split the heat into the heat transferred by the air and the heat trnasferred by the water:

Qa + Qw = 0

Qa = -Qw

The heat exchanged by the air is

Qa = Ga * Cpa * (tfin - ti)

And the heat exchanged by the water is:

Qw = Gw * Cpw * Δt

Replacing:

Ga * Cpa * (tfin - ti) = -Gw * Cpw * Δt

tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)

tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti

The G terms are mass flows, however we have volume flow of air.

With the gas state equation we calculate the mass:

p * V = m * R * T

m = (p * V) / (R * T)

55 °F = 515 °R

The gas constant for air is R = 53.35 (ft*lb)/(lbm* °R)

14.7 psi = 2117 lb/ft^2

m = (2117 * 5000) / (53.35 * 515) = 385 lbm

The mass flow is that much amount per minute

The mass flow of water is

11200 lbm/h = 186.7 lbm/min

Then:

tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F

Answer:

The answer is 535 nanometers.

Explanation:

[tex]1\ micrometer = 1\ \mu m = 1.0\times 10^{-6}\ m.[/tex]

and

[tex]1\ nanometer = 1\ nm = 1.0\times 10^{-9}\ m.[/tex],

so

[tex]1\ \mu m = 1.0 \times 10^{3}\ nm[/tex]

which means that

[tex]\lambda = 0.535\ \mu m = 535\ nm[/tex].

In fact we can say that the light is green, because its wavelength is in the range of 500 nm to 565 nm.

Answer:28 m

Explanation:

Given

Direction is [tex]58^{\circ}[/tex] North of east i.e. [tex]58 ^{\circ}[/tex] with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component [tex]=33sin58=27.98 m\approx 28 m[/tex]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

Explanation:

Given that,

Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]

Distance between charges, r = 1.81 m

(a) The electrostatic force that one charge exerts on the another is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]

[tex]F=9.21\times 10^{-8}\ N[/tex]

(b) As both charges are positively charged. So, the force of attraction between them is repulsive.

Answer:

Explanation:

So, lets say the runner stars from the position [tex]x_0[/tex]. Lets make this point the origin of a coordinate system in which the vector i points north.

[tex]x_0 = (0,0)[/tex]

Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:

[tex]x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)[/tex].

[tex]x_1 = (20 \ m,0)[/tex].

The, he runs 30 meters south

[tex]x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)[/tex]

[tex]x_2 = (-10 \ m,0)[/tex]

Finally, he runs 40 meter north

[tex]x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)[/tex]

[tex]x_3 = (30 \ m,0)[/tex].

This is our displacement vector. Now, the average speed will be:

[tex]\frac{distance}{time}[/tex].

The distance its the length of the displacement vector,

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

[tex]d=\sqrt{(30 \ m)^2+0^2}[/tex]

[tex]d=30 \ m[/tex]

So, the average speed its:

[tex]\frac{30 \ m }{30 \ s} = 1\frac{m}{s}[/tex].

The average velocity, instead, its:

[tex]\vec{v} = \frac{displacement}{time}[/tex]

[tex]\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}[/tex]

[tex]\vec{v} = (1 \ \frac{m}{s} ,\ 0)[/tex]

This is, 1 m/s north.