An electron falls through a distance d in a uniform electric field of magnitude E. Thereafter, the direction of the field is reversed (keeping its magnitude the same) and now a proton falls through the same distance. Compare, using quantitative reasoning, the time of fall in each case. Contrast this situation with that of objects falling freely under gravity.

Answer:

[tex]\vec{F} = -0.34\^x - 0.22\^y\\|\vec{F}| = -0.41~N[/tex]

Explanation:

The electric force between two point charges can be calculated by Coulomb's Law:

[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]

We have to calculate the distance between two points; (0,0) and (0.3 m, 0.2 m).

[tex]r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0.3)^2 + (0.2)^2} = 0.36~m[/tex]

Now we can apply Coulomb's Law

[tex]F = \frac{1}{4\pi\epsilon_0}\frac{(3\times 10^{-6})(-2\times 10^{-6})}{(0.36)^2} = -0.41~N[/tex]

The minus sign in front of the force means that the force is attractive.

The direction of the force can be calculated as follows:

[tex]F_x = F\cos(\theta)\\F_y = F\sin(\theta)[/tex]

where θ is the angle between F and the x-axis. This angle can be calculated by the triangle with edges 0.3 m, 0.2 m, and 0.36 m.  

So, sin(θ) = 0.2/0.36 = 0.55 and cos(θ) = 0.3/0.36 = 0.83.

Finally,

[tex]F_x = -0.41 \times 0.83 = -0.34~N\\F_y = -0.41 \times 0.55 = -0.22~N[/tex]

An electric force of approximately 0.0137 Newtons is exerted by the -2.00 μC charge on the 3.00 μC charge. The direction is attractive, implying that the force pulls the 3.00 μC charge towards the -2.00 μC charge.

The subject of this query pertains to the concept of electric force in Physics. Given the position coordinates and charge values, we can calculate the electric force between the two charges using Coulomb's Law, which states that the force between two charges is equal to the absolute value of the product of the charges, divided by the square of the distance between them, multiplied by the electrostatic constant (k = 8.99 x 10^9 N m²/C²).

First, determine the distance between the charges. Using the Pythagorean theorem, the distance is √(0.30² + 0.20²) = 0.36 m. Now plug the charge values (Q1=3.00 μC = 3.00 x 10^-6 C, Q2=-2.00 μC = -2.00 x 10^-6 C), and the distance (r=0.36 m) into Coulomb's Law (F=k|Q1*Q2|/r²).

The absolute value of the electric force would therefore be approximately |-2 x 8.99 x 10^9 N m²/C² x 3.00 x 10^-6 C x -2.00 x 10^-6 C / (0.36 m)²|, or about 0.0137 N (Newtons). Because force is a vector quantity, to find the direction of the force we need to consider the signs of the charges. Since they have opposite signs, the force is attractive, hence, the -2.00μC charge exerts a force towards itself on the +3.00 μC charge.

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Answer:

D = 3.45 mile

Explanation:

given,

time taken by the sound wave, t = 16.2 s

speed of sound, v = 343 m/s

speed of the light = 3 x 10⁸ m/s

distance where lightning stroke

 we know,

 distance = speed x time

  D = 343 x 16.2

  D = 5556.6 m

 1 m = 0.000621371 mile

5556.6 m = 5556.6 x 0.0006214

D = 3.45 mile

the distance lightning strike is 3.45 mile away from the observer.

To calculate the distance from a lightning stroke: use the time delay between seeing the lightning and hearing the thunder, and multiply it by the speed of sound (343 m/s). In this case, the lightning is approximately 5562.6 meters or about 3.46 miles away.

Your distance from a lightning stroke can be calculated by knowing the speed of sound and the time it takes for the sound from the lightning stroke to reach you. You said you hear the thunder sound 16.2 seconds after seeing the lightning. The speed of sound in air is 343 m/s. Therefore, the distance can be calculated using the formula: distance = speed * time.

In this case, distance = 343 m/s * 16.2s = 5562.6 meters or about 5.56 kilometers.

When this is converted to miles (since 1 mile is about 1609.34 meters), it is approximately 3.46 miles away. Therefore, based on the time difference between the flash and the thunder, you are about 5562.6 meters or 3.46 miles away from the lightning strike.

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Answer:

915.69549 m

306.1968 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 3.25 m/s²

g = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.25\times 230+0^2}\\\Rightarrow v=38.665\ m/s[/tex]

Height is given by

[tex]h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{38.665^2}{2\times 9.81}\\\Rightarrow h=76.1968\ m[/tex]

Total distance the canister has to travel is 230+76.1968 = 306.1968 m

Time to reach the max height

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{g}\\\Rightarrow t=\dfrac{0-38.665}{-9.81}\\\Rightarrow t=3.9413\ s[/tex]

Time taken to reach the ground is

[tex]s=ut+\frac{1}{2}gt^2\\\Rightarrow 306.1968=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{306.1968\times 2}{9.81}}\\\Rightarrow t=7.9\ s[/tex]

Total time taken would be 3.9413+7.9 = 11.8413 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=38.665\times 11.8413+\dfrac{1}{2}\times 3.25\times 11.8413^2\\\Rightarrow s=685.69549\ m[/tex]

The rocket is 685.69549+230 = 915.69549 m from the launchpad.

The total distance is 230+76.1968 = 306.1968 m

Answer:

No, the ride will not be safe according to the standards mentioned in the question because the value of momentum exceeds.

Explanation:

Given:

Now, we find the momentum of the car using the given mass and speed:

[tex]p'=m.v[/tex]

[tex]p'=780\times 3[/tex]

[tex]p'=2340\ kg.m.s^{-1}[/tex]

Therefore the ride will not be safe according to the standards mentioned in the question because the value of momentum exceeds.

Answer:

v = 5.05m/s

Explanation:

H = 1.3m

initial velocity = 0

final velocity = v = ?

g =9.8 m/s^2

we apply the conservation of energy; all potential energy is comletely converting to kinetic energy

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

mass is same; it cancels out

[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]

v = 5.05m/s

Answer:

the final linear velocity is 3.56931 m/s

Explanation:

the solution is in the attached Word file

Answer:

[tex]L_f=26.8108 ft[/tex]

Part B:

For Final Reduction

[tex]v_f=48.5436ft/min[/tex]

Explanation:

Part A:

At each step 0.8 (100-20)% of thickness is left

Final Thickness t_f:

[tex]t_f=(0.80)(0.80)(0.80)*3\\t_f=1.536 in[/tex]

Width increases by 0.03 in each step so (100+3)%=1.03

Final Width w_f:

[tex]w_f=(1.03)(1.03)(1.03)*12\\w_f=13.1127 in[/tex]

Conservation of volume:

[tex]t_ow_oL_o=t_fw_fL_f\\L_f=\frac{t_ow_oL_o}{t_fw_f} \\L_f=\frac{3*12*(15*12)}{1.536*13.1127}\\L_f=321.730 in\\L_f=26.8108 ft[/tex]

Part B:

[tex]t_ow_ov_o=t_fw_fv_f[/tex]

At First reduction exit Velocity:

[tex]v_f=\frac{t_ow_oL_o}{t_fw_f} \\v_f=\frac{3*12*(40)}{0.8*3*1.03*12}\\v_f=48.5439ft/min[/tex]

At 2nd Reduction:

[tex]v_f=\frac{0.8*3*1.03*12*40}{0.8^2*3*1.03^2*12}\\v_f=48.5436ft/min[/tex]

For Final Reduction:

[tex]v_f=\frac{0.8^2*3*1.03^2*12*40}{0.8^3*3*1.03^3*12}\\v_f=48.5436ft/min[/tex]

Answer:

The force that the magnetic field of the current exerts on the electron is [tex]2.30\times10^{-19}\ N[/tex]

Explanation:

Given that,

Distance = 4.40 cm

Speed [tex]v= 6.10\times10^{4}\ m/s[/tex]

Suppose a long, straight wire carries a current of 5.20 . An electron is traveling in the vicinity of the wire.

We need to calculate the magnetic field of the current exerts on the electron

Using formula of force

[tex]F=qv\times B[/tex]

[tex]F=qv\times\dfrac{\mu_{0}I}{2\pi r}[/tex]

Put the value into the formula

[tex]F=1.6\times10^{-19}\times6.10\times10^{4}\times\dfrac{4\pi\times10^{-7}\times5.20}{2\pi\times4.40\times10^{-2}}[/tex]

[tex]F=2.30\times10^{-19}\ N[/tex]

Hence, The force that the magnetic field of the current exerts on the electron is [tex]2.30\times10^{-19}\ N[/tex]

Answer:

1.29649

488.08706 nm

[tex]6.14644\times 10^{14}\ Hz[/tex]

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

[tex]\lambda_0[/tex] = 632.8 nm

From Snell's law we have the relation

[tex]n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649[/tex]

Refractive index of the solution is 1.29649

Wavelength is given by

[tex]\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm[/tex]

The wavelength of the solution is 488.08706 nm

Frequency is given by

[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz[/tex]

The frequency is [tex]6.14644\times 10^{14}\ Hz[/tex]

[tex]v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s[/tex]

The speed in the solution is 231715700.28346 m/s

The index of refraction can be found using Snell's law. The wavelength of red light within the solution can be calculated from the index and its frequency remains constant. The speed of light in the solution is the product of its frequency and the calculated wavelength.

Solution to the Laser Beam Question

To answer the student's question regarding refraction of a laser beam in a corn syrup solution, we can use Snell's law, which relates the incident angle, refracted angle, and indices of refraction of the two media. We can also calculate the wavelength, frequency, and speed of light within the solution using the principles of wave optics.

(a) According to Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n2 is the index of refraction of the corn syrup solution, θ1 is the incident angle, 90° - 33.0° = 57.0°, and θ2 is the refracted angle, 90° - 24.84° = 65.16°. Since the laser light is passing from air (n1=1) to the solution, we have 1 * sin(57.0°) = n2 * sin(65.16°). Solving for n2 gives us the index of refraction of the syrup solution.

(b) The wavelength of light in the solution is given by λ' = λ0/n, where λ0 is the wavelength in a vacuum, and n is the index of refraction. Inserting the red light's wavelength (632.8 nm) in vacuum and the obtained index of refraction will yield the wavelength in the solution.

(c) The frequency of light does not change when it enters another medium, so we use the vacuum frequency calculated by f = c/λ0, and c is the speed of light in a vacuum.

(d) The speed of light in the solution, v, can be found using the equation v = f * λ', which uses the frequency found in part (c) and the wavelength in the solution from part (b).

Explanation:

a)

Sum of moments = 0 (Equilibrium)

T . cos (Q)*L = m*g*L/2

[tex]cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}[/tex]

[tex]T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2[/tex]

[tex]T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }[/tex]

b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!

c)

[tex]T = \frac{1.962}{\sqrt{(2)^2 - L^2} }[/tex]

The rotational equilibrium condition allows finding the responses for the tension of the rope are:

   A) T = 0.53 N

   B) If the rope shortens the tension increase..

   C) T = 0.57 N

Newton's second law for rotational motion gives a relationship between the torque, the moment of inertia and the angular acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition.

                  Σ τ = 0

Torque is defined as the vector product of the force and the distance, its modulus is:

                τ = F rsin θ  

where τ is the torque, F the force, r the distance and tea the angle between the force and the distance, it is a product (r sin θ ) it is called the perpendicular distance or arm.

In the attached we have a free body diagram of the system, let's apply the equilibrium condition,

Let's use trigonometry to descompose the force.  

                cos θ = [tex]\frac{T_x}{T}[/tex]  

               sin  θ = [tex]\frac{T_y}{T}[/tex]  

               Tₓ = T cos θ

               [tex]T_y[/tex] = T sin  θ

They indicate that the length of the bar is x = 1 m and the length of the cable is L = 2.5 m, let's find the angle

             cos  θ = [tex]\frac{x}{L}[/tex]  

              θ = cos⁻¹ [tex]\frac{x}{L}[/tex]  

              θ = cos⁻¹ [tex]\frac{1}{2.5}[/tex]

              θ = 66.4º

A) let's set our pivot point at the junction with the wall and the anti-clockwise direction of rotation is positive.

                 W [tex]\frac{L}{2}[/tex]  - Ty L = 0

                 W [tex]\frac{L}{2}[/tex]  = (T sin 66.4) L

                 [tex]T = \frac{mg}{s sin 66.4} \\ \\T = \frac{0.1 \ 9.8}{2 \ sin66.4}[/tex]  

                 T = 0.53 N

B) how the tension changes as the length of the string changes.

               T = [tex]\frac{mg}{2 sin \theta}[/tex]  

we can see that the change of the tension occurs by changing the value of the sine function.

           sin θ = [tex]\frac{y}{L_o}[/tex]

Let's use the Pythagorean theorem to find the opposite leg.

         L² = x² + y²

         y = [tex]\sqrt{L^2 - x^2 }[/tex] = [tex]L \ \sqrt {1^2 + (\frac{x}{L})^2 }[/tex]

Let's substitute.

         sin θ =  [tex]\sqrt{1 - \frac{x^2}{L^2} }[/tex]

if we use a binomial expansion.

          [tex](1 - a) ^{0.5} = 1 - \frac{1}{2} a + ...[/tex]

Let's substitute.

            sin  θ  = 1 -  [tex]\frac{1}{2} \ \frac{x}{L}[/tex]    

We can see that when the value of the length decreases the value of the sine decreases and this term is in the denominator of the expression of the tension, therefore the tension must increase.

C) the length of the rope is L = 2 m

           sin θ =  [tex]\sqrt{1 - (\frac{1}{2})^2 }[/tex]  

           sin θ = 0.866

           T = [tex]\frac{mg}{2sin \theta}[/tex]  

            T = [tex]\frac{0.1 \ 9.8}{2 \ 0.866}[/tex]  

            T =0.57 N

In conclusion, using the rotational equilibrium condition we can find the answers for the tension of the rope are:

   A) T = 0.53 N

   B) If the rope shortens the tension increase.

   C) T = 0.57 N

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Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are  in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

To find the equivalent resistance Req between terminals a and b, we can follow a series of steps to simplify the circuit. The final equivalent resistance can be calculated by adding the resistors in series. With terminals c and d open, Req is 12.22 ohms, and with terminals c and d shorted, Req is 22.00 ohms.

To find the equivalent resistance between terminals a and b, we need to consider the circuit shown in Figure 10.15. To simplify the circuit, we can apply a series of steps to reduce it to a single equivalent resistance. Step 1 involves reducing resistors R3 and R4 in series. Step 2 involves reducing resistors R2 and the equivalent resistance R34 in parallel. Finally, Step 3 involves reducing resistor R1 and the equivalent resistance R234 in series. The final equivalent resistance Req can be calculated by adding the resistors in series.

Following these steps, the equivalent resistance Req between terminals a and b with terminals c and d open is 12.22 ohms. With terminals c and d shorted together, the equivalent resistance Req is 22.00 ohms.

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Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy  = 1 / 7

λy / λx = 7 / 1

The ratio of wavelengths is λ(y)/λ(x) = 1/49.

The energy of a wave is directly proportional to the square of the amplitude of the wave.

   E ∝ [tex]A^{2}[/tex] , where E is the energy of the wave and A is the apmlitude

We also know that enegy

E = hc/λ

E ∝ 1/λ

According to the question:

Let wave X has energy [tex]E_x[/tex] and amplitude [tex]A_x[/tex]

and wave Y has energy [tex]E_y[/tex] and amplitude [tex]A_y[/tex]

[tex]\frac{E_x}{E_y}=\frac{A_x^2}{A_y^2} \\\\\frac{E_x}{E_y}= \frac{(A_y/7)^2}{A_y}\\\\\frac{E_x}{E_y}=\frac{1}{49}[/tex] since it is given that [tex]A_x=\frac{1}{7}A_y[/tex]

{1/λ(x)} / {1/λ(y)} = 1/49

λ(y)/λ(x) = 1/49 is the ratio of the wavelengths.

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Final answer:

The question pertains to the physics discipline, dealing with calculating the tension in supporting cables using principles of static equilibrium and considering the load's weight and the angles of attachment.

Explanation:

The question involves determining the tension in cables that support a structure or an object. This type of problem is common in physics, specifically in the area of mechanics, and it involves understanding forces and how they are distributed within a system. When multiple cables support a load, tensions in each cable can be found using principles of static equilibrium, where the sum of forces in each direction (horizontal and vertical) equals zero. In addition to the weight of the supported object, angles at which cables are attached can play a crucial role in determining individual tensions.

Answer:

d. 500 m/s

Explanation:

Momentum: This is the product of the mass of a body to its velocity, The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv....................... equation 1

Where M = momentum, m = mass, v = velocity

deduced from the question,

Momentum of the car = momentum of the barrier.

MV = mv ............................. Equation 1

Where M = mass of the car, V = velocity of the car, m = mass of the barrier, v = velocity of the barrier.

making v the subject of the equation,

v = MV/m........................ Equation 2

Given: M = 1000 kg, V = 10 m/s, m = 20 kg.

Substitute into equation 2

v = 1000(10)/20

v = 500 m/s.

Hence the speed of the barrier = 500 m/s

The right option is d. 500 m/s

Answer:

0.26315 s

Explanation:

The frequency of the ball tied to a string system is 3.8 rev/s.

That means in one second the ball will complete 3.8 revolutions.

The time period will be the reciprocal of this frequency

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.8}\\\Rightarrow T=0.26315\ s[/tex]

The time period is 0.26315 s

It can be also solved in the following way

[tex]1\ s=3.8\ rev\\\Rightarrow 1\ rev=\dfrac{1}{3.8}\ s\\\Rightarrow 1\ rev=0.26315\ s[/tex]

The time period is 0.26315 s

Final answer:

The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.

Explanation:

When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:

F = k * (Q1 * Q2) / r^2

where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

We know that the gravitational force is equal to the weight of the paper:

Fg = m * g

where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).

At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:

F = Fg

k * (Q1 * Q2) / r^2 = m * g

We can solve this equation for the magnitude of the charge Q1 or Q2:

Q1 * Q2 = (m * g * r^2) / k

Substituting the known values, we get:

Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)

Q1 * Q2 = 1.383 x 10^-12 C^2

To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:

Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C

The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.

To warm the cold air inhaled during a breath to body temperature requires about 35.05J of energy or heat. This results in a total loss of around 44.16 KJ of heat due to respiration per hour.

This is a question about the heat exchange between the human body and the surrounding air when breathing, in cold weather conditions. We can solve it using the concepts of heat transfer and specific heat capacity.

A. To find the amount of heat necessary to warm the cold air to body temperature, we first need to determine the change in temperature. This would be (37-(-19)) = 56°C. The mass of the air breathed in with each breath can be calculated as 0.470L * 1.3g/L = 0.611g. Converting this to kg, we get 0.000611 Kg. The quantity of heat, Q, can be found using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, we get Q = 0.000611 Kg * 1020 J/Kg°C * 56°C = 35.05J.

B. The total amount of heat lost per hour due to respiration can be calculated by first finding the heat lost per minute: Q per minute = Q per breath * respiration rate = 35.05J/breath * 21 breaths/minute = 736.05 J/minute. Converting this to an hourly rate gives us 736.05 J/minute * 60 minutes/hour = 44,163J/hour or about 44.16 KJ/hour. So, about 44.16 KJ of heat is lost via respiration each hour.

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The amount of heat needed to warm to internal body temperature the air exchanged with each breath on a cold day is 74.3 J per breath. The total heat loss per hour for a respiration rate of 21.0 breaths per minute is approximately 93.6 kJ/hour.

To calculate the amount of heat needed to warm the air inhaled during breathing to the internal body temperature, we can use the formula:

Q = mcΔT

Where:

The mass of 1.0 L of air at 1.3 g/L is 0.0013 kg. The change in temperature needed to warm air from -19°C to 37°C is 56°C or 56 K since the size of 1 degree on both scales is the same. Thus:

Q = 0.0013 kg * 1020 J/kg·K * 56 K

Q = 74.3 J per breath.

To find the heat lost per hour at a respiration rate of 21.0 breaths per minute, we do:

Heat loss per hour = Q * number of breaths per hour = 74.3 J * (21 breaths/min * 60 min/hour)

Heat loss per hour = 74.3 J * 1260 breaths/hour = 93618 J/hour, or approximately 93.6 kJ/hour.

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Answer:

The half-life of A is 17.1 days.

Explanation:

Hi there!

The half-life of B is 1.73 days.

Let´s write the elapsed time (3 days) in terms of half-lives of B:

1.37 days = 1 half-life B

3 days = (3 days · 1 half-life B / 1.37 days) = 2.19 half-lives B.

After 3 days, the amount of A in terms of B is the following:

A = 4.04 B

The amount of B after 3 days can be expressed in terms of the initial amount of B (B0) and the number of half-lives (n):

B after n half-lives = B0 / 2ⁿ

Then after 2.19 half-lives:

B = B0 /2^(2.19)

In the same way, the amount of A can also be expressed in terms of the initial amount and the number of half-lives:

A = A0 / 2ⁿ

Replacing A and B in the equation:

A = 4.04 B

A0 / 2ⁿ = 4.04 · B0 / 2^(2.19)

Since A0 = B0

A0 / 2ⁿ = 4.04 · A0 / 2^(2.19)

Dividing by A0:

1/2ⁿ = 4.04 / 2^(2.19)

Multipliying by 2ⁿ and dividing by  4.04 / 2^(2.19):

2^(2.19) / 4.04 = 2ⁿ

Apply ln to both sides of the equation:

ln( 2^(2.19) / 4.04) = n ln(2)

n = ln( 2^(2.19) / 4.04) / ln(2)

n = 0.1756

Then, if 3 days is 0.1756 half-lives of A, 1 half-life of A will be:

1 half-life ·(3 days / 0.1756 half-lives) = 17.1 days

The half-life of A is 17.1 days.

Answer:

i) They are equal . ii) It is thousands of times greater for the electron.

Explanation:

i) By definition, the electric field is the electric force per unit charge. If the field is the same, the force will depend on the value of the charge under the influence of the field.

As the magnitude of  the charge of the electron and the proton are the same, we conclude that the electric force on both must be equal in magnitude.

ii) The acceleration on both particles must meet the Newton´s 2nd Law, so, if the forces are equal in magnitude (neglecting any other external interaction), the acceleration will only depend on the mass of both particles, according this general expression:

a = F/m

As the mass of the electron is approximately two thousands times smaller than the proton´s, it concludes that the acceleration on the electron must be thousands of times greater for the electron.

The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. The magnitudes of their accelerations are equal.

(i) The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. This is because the electron has a much smaller mass compared to the proton, and the electric force depends on the charge and mass of the particle.

(ii) The magnitudes of their accelerations are equal. The acceleration of a particle in an electric field depends only on the magnitude of the electric field and the mass of the particle. Since both the electron and proton experience the same electric field, their accelerations will be the same.

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Answer:

0.03981 W/m²

Explanation:

I = Sound intensity

[tex]\beta[/tex] = Intensity level = 106 dB

[tex]I_0[/tex] = Threshold of human hearing = [tex]10^{-12}\ W/m^2[/tex]

Intensity of sound is given by

[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow 106=10log\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{106}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{106}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow I=10^{\dfrac{106}{10}}\times 10^{-12}\\\Rightarrow I=10^{-1.4}\\\Rightarrow I=0.03981\ W/m^2[/tex]

The sound intensity is 0.03981 W/m²

The sound intensity is approximately 1.00 x 10^-4 W/m^2.

The sound intensity can be calculated using the formula:

I = I0 * 10^(dB/10)

where I is the sound intensity, I0 is the threshold of human hearing (given as 1.00 x 10^-12 W/m^2), and dB is the decibel level above the threshold. In this case, the decibel level is 106 dB. Plugging in the given values into the formula:

I = (1.00 x 10^-12 W/m^2) * 10^(106/10)

Simplifying the equation:

I ≈ 1.00 x 10^-4 W/m^2

The sound intensity is approximately 1.00 x 10^-4 W/m^2.

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The can of soft drink in a refrigerator is modelled as a closed system because while heat energy is exchanged with its surroundings, no matter is exchanged. An open system, in contrast, allows both energy and matter to be exchanged.

In the scenario where a can of soft drink is placed into a refrigerator to cool, you would model the can of soft drink as a closed system. A closed system is one in which energy can be exchanged with its surroundings, but matter cannot. In this case, the can of soft drink is transferring heat energy to its surroundings, i.e., the refrigerator, until equilibrium is achieved, but the soda itself remains within the can - no matter is exchanged.

An open system, on the other hand, allows both energy and matter to be exchanged with its environment. The dissolving of CO2 in soft drinks, as mentioned in the reference, is an example of an open system, where matter (CO2) is able to escape from the soft drink when the can or bottle is opened.

It's important to note that while practical real-world systems are ultimately open due to unavoidable interactions with the surroundings, we often model systems as closed (or even isolated) in order to simplify the analysis.

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The correct answer is that the can of soft drink would be modeled as a closed system.

In thermodynamics, a closed system is defined as a system that can exchange energy (such as heat) with its surroundings, but not matter. This means that while the can of soft drink can absorb or release heat to the refrigerator (its surroundings), no matter (like the soft drink itself) is exchanged between the can and the refrigerator.

Therefore, when the can of soft drink is placed in the refrigerator to cool, it is appropriately modeled as a closed system because only energy in the form of heat is transferred between the can and the refrigerator, and there is no transfer of matter.

Final answer:

For the return flight, the bird's average velocity is 6.38 m/s, while for the whole episode, the average velocity is 0 m/s.

Explanation:

a) To find the bird's average velocity for the return flight, we need to calculate the displacement and divide it by the time taken. The displacement is the distance between the release point and the nest, which is 5220 km. The time taken is 13.3 days, which can be converted to seconds by multiplying by 24 (hours in a day) and 60 (minutes in an hour). Therefore, the average velocity is:

Average velocity = displacement / time = 5220 km / (13.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 6.38 m/s

b) To find the bird's average velocity for the whole episode, we need to calculate the total displacement and divide it by the total time taken. The total displacement is 0 km, as the bird returns to its nest. The total time taken is the time taken for the return flight plus the time taken for the outward flight, which is 2 * 13.3 days. Therefore, the average velocity is:

Average velocity = total displacement / total time = 0 km / (2 * 13.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 0 m/s

Answer:

3333.33 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to its volume.

The unit of density is kg/m³.

From Archimedes principle,

R.d = W/U = D/D'

Where R.d = relative density, W = weight of the object in air, u = upthrust in water, D = Density of the object, D' = Density of water.

W/U = D/D'

making D the subject of the equation

D = D'(W/U).......................... Equation 1

Given: W = 5.0 N, U = 5.0 -3.5 = 1.5 N, D' = 1000 kg/m³

Note: U = lost in weight = weight in air - weight in water

Substitute into equation 1

D = 1000(5/1.5)

D = 3333.33 kg/m³

Thus the density of the object = 3333.33 kg/m³

The first step is to calculate the buoyant force that the water exerts on the solid object. The buoyant force can be found by subtracting the scale reading from the gravitational force, or 5.00N - 3.50N = 1.50N.

Next, we find the volume of the water displaced by the solid. The buoyant force is equal to the weight of the fluid displaced, so we can use the formula F = ρf * V * g, where ρf is the fluid density, V is the water volume displaced, g is the acceleration due to gravity and F is the buoyant force.

We're given that the density of water is 9.8 * 1000 N/m³ and we've just calculated the buoyant force. Thus, we have 1.50N = 9.8 * 1000 * V * 9.81, which simplifies to V = 1.5 / (9.8 * 1000) = 0.0001530612244897959 m³.

Now we calculate the object's mass using the equation, m = F / g, where F is gravitational force and g is acceleration due to gravity. Substituting the given values, we get m = 5.00 / 9.81 = 0.509683995922528 kg.

Finally, we find the density of the object using the formula ρ = m / V, where m is the mass and V is the volume. Substituting the values calculated earlier, we get ρ = 0.509683995922528 / 0.0001530612244897959 = 3329.935440027183 kg/m³.

Therefore, the density of the object is approximately 3330 kg/m³.

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Answer:

At twice the distance, the electric potential is V/2.

Explanation:

The electric potential at a certain distance d from a point charge q can be represented by V:

V = kq/d .....1

Where q = charge ,d = distance between them, k = coulomb's constant.

When the distance is doubled d1 = 2d

Let V1 represent the electric potential after the distance have been doubled.

V1 = kq/2d = (kq/d)/2

V1 = V/2

Therefore, at twice the distance the electric potential is halved V1 = V/2

The electric potential (V) decreases as you increase your distance from the point charge. Therefore, at twice the distance from a charge, the electric potential will be halved, becoming V/2.

The electric potential (V) at a certain distance from a point charge, according to Coulomb's Law, is directly related to the charge and inversely related to the distance from it. Therefore, if we double the distance, the electric potential will be half as much. So, the electric potential at twice the distance from the point charge is V/2.

Think of it like this, if you're twice as far from the point charge, the effect of the charge (which in this case is represented by the electric potential) is lessened.

Therefore, the potential decreases as you increase your distance from the charge. Hence, at twice the distance from a charge, the electric potential will be halved, i.e., V/2.

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Answer:

D.frequency.

Explanation:

The number of times a wave passes through a particular point in a time of 1 second is called the frequency.

Amplitude is the maximum height the wave reaches from the reference axis of the wave

Wavelength is the distance between the two upper (crest) or lower (troughs) points of a wave.

Crest is the top most part of a wave.

Hence, the question is referring to frequency.

Answer: The correct explanation is 2.

Explanation: The warm air is less dense (it expands) and thus it is lighter than the cold air so it will rise up to the floor. Therefore, when you place the heater on the floor it will warm the cold air which would then rise and be replaced by more cold air which would again get warm and rise and so on until the room is heated. This means that the correct explanation is 2.

On the other hand, if you put the heater at the ceiling, it will warm the cold air near the ceiling which would stay up there (it is lighter than the cold air under it). This means that the only way for the heat to spread from this ceiling level warm air to the lower levels is via conduction which is slow.  

Final answer:

The heater should be placed near the floor due to convection, where warm air rises after being heated, circulates, and evenly distributes throughout the room as it cools and descends. Therefore, option 2 is the correct answer.

Explanation:

Proper Placement of a Heater in a Room

When considering the placement of a heater in a room in the context of convection, the better option is to place the heater near the floor. This is because, as per the principles of convection, warm air will rise after being heated by the heater. The now-warm air, having become less dense, will rise to the ceiling, creating a convective loop that circulates the warm air throughout the room. As the air cools down near the ceiling and outside walls, it contracts, becoming denser, and subsequently sinks back to the floor, where it will be heated again by the heater. This cycle continues, leading to an efficient distribution of heat throughout the room. Therefore, option 2 is correct: placing the heater near the floor allows it to warm the air close to it, after which the warm air rises and is replaced by cool air that the heater warms again, maintaining a continuous cycle of heating.

Explanation:

According to the energy conservation,

          [tex]F_{centripetal} = F_{electric}[/tex]

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

           [tex]v^{2} = \frac{kq^{2}r}{d^{2}m}[/tex]

                 = [tex]\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}[/tex]

                = [tex]8.430 \times 10^{10} m^{2}/s^{2}[/tex]

             v = [tex]\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}[/tex]

                = [tex]2.903 \times 10^{5} m/s[/tex]

Formula for distance from the orbit is as follows.

               S = [tex]2 \pi r[/tex]

                  = [tex]2 \times 3.14 \times 0.75 \times 10^{-9} m[/tex]

                  = [tex]4.71 \times 10^{-9} m[/tex]

Now, relation between time and distance is as follows.

                T = [tex]\frac{S}{v}[/tex]

       [tex]\frac{1}{f} = \frac{S}{v}[/tex]

or,           f = [tex]\frac{v}{S}[/tex]          

                = [tex]\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}[/tex]      

                = [tex]6.164 \times 10^{13} Hz[/tex]

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is [tex]6.164 \times 10^{13} Hz[/tex].

Answer:

The pressure is expected to decrease because as we move up in the atmosphere the is lesser mass of atmosphere weight above us.

Explanation:

The atmospheric pressure is the the force exerted by the layers of the atmosphere above a point. Being a fluid and due to the gravity of earth there is a pressure due to the weight of the atmospheric columns above us.

So, at a height h above the earth surface in the atmosphere the pressure will decrease and this can be calculated as:

[tex]P_h=101325\times (1-2.25577\times 10^{-5}\times h)^{5.25588}[/tex]

putting h=9.6 meter

[tex]P_h=101209.7268\ Pa[/tex]

At an elevation of 9.6 km, the pressure is approximately 254 millibars.

The pressure is inversely proportional to elevation, higher the elevation lower will be the pressure while on the other hand, lower the elevation higher will be the pressure. At an elevation of 10 km pressure drops to 265 millibars then the elevation of 9.6 km, the pressure is approximately 254 millibars.

The atmospheric pressure at sea level is about 1,014 millibars which is equal to the pressure of 14.7 pounds/inch2 so we can conclude that at the elevation of 9.6 km, the pressure is approximately 254 millibars.

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Answer:

9.6609 rad/s

10.143945 m/s

Explanation:

I = Moment of inertia = 3.75 kgm²

K = Kinetic energy = 175 J

r = Radius = 1.05 m

Kinetic energy is given by

[tex]K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s[/tex]

The angular velocity of the leg is 9.6609 rad/s

Velocity is given by

[tex]v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s[/tex]

The velocity of the tip of the punters shoe is 10.143945 m/s

Answer:

It is constant.

Explanation:

As we know that the electric field is a change in the electric potential so,

E = -ΔV

E = 0 ∵ V = constant

As we know that the electric field is zero. Which means that the electric potential is constant and it's not changing which results in the zero electric field.

In a region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero.

In a certain region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero. This is because electric potential is determined by the distribution of charges and their distances from the region in consideration. While the electric field measures the force experienced by a charge, the electric potential is related to the work required to move a charge in a region with an electric field.

Therefore, the correct answer is (e) None of those answers is necessarily true.

Answer:

v1 = 377.98 m/s

Explanation:

m = 5 Kg

v0 = 176 m/s

v0x = v0*Cos 32° = 176 m/s*Cos 32° = 149.256 m/s

m1 = 2 Kg

m2 = 3 Kg

t = 4.1 s

g = 9.81 m/s²

Before  the explosion

pix = m*v0x = 5 Kg*149.256 m/s = 746.282 Kgm/s

piy = 0

After the explosion

pfx = m1*v1x

knowing that pix = pfx

we have

746.282 = 2*v1x

v1x = 373.14 m/s

v2y = g*t

pfy = m1*v1y + m2*v2y

pfy = 2*v1y + 3*(9.81*4.1)

pfy = 2*v1y + 120.663

knowing that piy = pfy = 0

we have

0 = 2*v1y + 120.663

v1y = 60.33 m/s

Finally we apply

v1 = √(v1x² + v1y²)

v1 = √(373.14² + 60.33²)

v1 = 377.98 m/s

Final answer:

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we use the principle of conservation of momentum. By setting the momentum before the explosion equal to the momentum after the explosion, we can solve for the velocity of the 2 kg fragment. The magnitude of the velocity is approximately 56.8 m/s.

Explanation:

Projectile Motion

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we need to use the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion is equal to the total momentum after the explosion.

Before the explosion, the momentum of the projectile can be calculated using the formula:

momentum = mass * velocity

After the explosion, the momentum of the 2 kg fragment can be calculated as:

momentum = mass * velocity

Setting the two equations equal to each other, we can solve for the velocity of the 2 kg fragment.

Plugging in the given values:

Initial velocity of the projectile = 176 m/s
Initial angle = 32°
Mass of the 2 kg fragment = 2 kg
Mass of the 3 kg fragment = 3 kg
Time after explosion = 4.1 s
Acceleration due to gravity = 9.81 m/s^2

Solving the equation, we find that the magnitude of the velocity of the 2 kg fragment immediately after the explosion is approximately 56.8 m/s.