The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation.

Light takes 320 minutes to reach Earth from Pluto

Explanation:

We can solve this problem by applying the rule of three. We can do as follows:

- We call [tex]x[/tex] the distance between the Sun and the Earth

- Then the distance between the Earth and Pluto is 40 times this distance, so [tex]40x[/tex]

- Light takes about 8 minutes to cover the distance x between Sun and Earth

Therefore, calling T the time that light takes to cover distance between Earth and Pluto (40x), we can write:

[tex]\frac{x}{8}=\frac{40x}{T}[/tex]

And solving for T,

[tex]T=\frac{8\cdot 40 x}{x}=320 min[/tex]

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Answer:

[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.

Explanation:

Given:

Time taken by the canister to hit the ground:

using equation of motion

[tex]h=u_y.t+\frac{1}{2} g.t^2[/tex]

where:

[tex]u_y=[/tex] initial vertical velocity of the canister = 0 (since the the object is dropped from a horizontally moving plane)

[tex]t=[/tex] time taken to hit the ground

[tex]90=0+0.5\times 9.8\times t^2[/tex]

[tex]t=4.2857\ s[/tex]

Now the horizontal distance travelled by the canister after dropping:

[tex]s=v_x\times t[/tex]

[tex]s=64\times 4.2857[/tex]

[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.

Answer:

R = 4.18 * 10^8ohms

Explanation:

R=resistance in ohms=?

ρ=resistivity of material in ohms meters = 1.68*10^-8 oh ohm meters

L= length of the object (m) = 4.4m

A = cross-sectional area of the object in square meters (m^2)= πr^2

r = (1.5mm/1000)/2= 0.00075m

A=π*(0.00075)^2 = 1.76714586764426*10^-6

Approximately, A = 1.767m^2

R = ρL/A= (1.68*10^8Ω⋅m) * (4.4m)/(1.767m^2)

R = 418336162.988115 ohms

Approximately, R = 4.18 * 10^8ohms.

The resistivity of copper will be "4.18 × 10⁸ Ω (ohms)".

According to the question,

Resistivity of ohms, ρ = 1.68 × 10⁻⁸ Ω.m

Object's length, L = 4.4 m

Radius, r = [tex]\frac{\frac{1.5}{1000} }{2}[/tex]

               = 0.00075 m

We know,

The cross-sectional area,

A = πr²

By substituting the values,

  = π × (0.00075)²

  = 1.767 × 10⁻⁶

  = 1.767 m²

hence,

The resistivity will be:

→ R = [tex]\frac{\rho L}{A}[/tex]

      = [tex]\frac{1.68\times 10^{-8}\times 4.4}{1.767}[/tex]

      = 4183316162.9 Ω or,

      = 4.18 × 10⁸ Ω  

Thus the above answer is correct.

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Answer:

3.8 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of  the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) =  9.564 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = (  9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 9.564 × 10⁻¹³ = 3.8 × 10 ⁻¹⁴ m

Answer:

13.01 m/s²

Explanation:

The period of a simple pendulum is given as

T = 2π√(L/g) .......................... Equation 1

Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.

Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14

Substitute into equation 1

1.28 = (2×3.14)√(0.54/g)

1.28 = 6.28√(0.54/g)

Making g the subject of the equation,

√(0.54/g) = 1.28/6.28

√(0.54/g) = 0.2038

0.54/g = (0.2038)²

0.54/g = 0.0415

g = 0.54/0.0415

g = 13.01 m/s²

Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

In this exercise we have to use the pressure knowledge to calculate the velocity, temperature and density so we have:

The variation of temperature and pressure is given by the formula of:

[tex](P_0/P) = (T_0/T)^{(k/(k-1))}[/tex]

From the formula given above we can identify:

Solving the formula  for temperature we find:

[tex](101000/101030) = (303.15/T)^{(1.4/(1.4-1))}\\0.9990703 =(303.15/T)^{(3.5)}\\T = 303.1758 K\\\Delta T = T - T_0 = 303.1758 - 303.15 = 0.0258 K[/tex]

They are using the formula for density is:

[tex]\rho_0 = 101000/(287 * 303.15) = 1.1608655 kg/m^3\\\rho = P/RT\\\rho = 101030/(287*303.1758) = 1.1611115 kg/m^3\\\Delta \rho = 2.4 * 10^{-4} kg/m^3[/tex]

Calculating the speed we find that:

[tex]c_0 = \sqrt{kRT_0} = 349.00669 m/s\\c = \sqrt{kRT} = 349.0215415 m/s\\\Delta c = 0.0148 m/s[/tex]

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Answer:

The speed the glow-worm would have accelerated after 10 years if released into free space and assumed to live is 3.09 X 10³⁴ m/s

Explanation:

Energy associated with photon of lights, is given as

E = hc/λ

Where;

h is Planck's constant =  6.626 x 10⁻³⁴ m2 kg/s

C is the speed of light = ?

λ is the light's wavelength = 650nm = 650 X10⁻⁹ m

Also Energy = Power X time

Time given = 10 years

Time in seconds = 10 yrs X 365 days X 24 hrs X 60mins X 60 secs

Time in seconds = 315360000 seconds

P*t = hc/λ

c = (P*t*λ)/h

c = (0.1 X 315360000 X 650 X 10⁻⁹)/(6.626 x 10⁻³⁴)

c = 3.09 X 10³⁴ m/s

The speed the glow-worm would have accelerated after 10 y if released into free space and assumed to live is 3.09 X 10³⁴ m/s

Final answer:

The seed takes approximately 0.47 seconds to reach its maximum height when launched at 90° with the highest possible initial speed of 4.6 m/s.

Explanation:

When a seed is launched with an initial speed of 4.6 m/s and at an angle of 90°, the vertical component of the initial velocity is 4.6 m/s. The time it takes for a projectile to reach its maximum height can be calculated using the equation for vertical motion:

t = (Vf - Vi) / g

Where:

t is the time

Vf is the final velocity (which is 0 m/s at the maximum height)

Vi is the initial velocity

g is the acceleration due to gravity (9.8 m/s²)

Using the given information, we can calculate the time it takes for the seed to reach its maximum height:

t = (0 - 4.6) / -9.8 = 0.469 s

Therefore, the seed takes approximately 0.47 seconds to reach its maximum height.

The statement "The magnitude of the physical quantity of electric current \( I \) in a wire equals the magnitude of the electric charge q that passes through a cross-section of the wire divided by the time interval [tex]\( \Delta t \)[/tex] needed for that charge to pass" could be an operational definition of electric current.

This definition precisely defines electric current as the rate of flow of electric charge through a conductor over a specific time period.

By quantifying the amount of charge passing through a cross-section of the wire in a given time interval, this definition provides a measurable and practical way to understand and quantify electric current.

It aligns with the fundamental concept that electric current represents the flow of charge, making it a suitable operational definition in the context of electrical systems and circuits.

Use Newton's second law to find the mass in kilograms, and finally convert it to pounds-mass, which is approximately 14.52 pounds-mass.

The question asks us to calculate the mass of the spacecraft given the thrust of the first booster stage and the acceleration of the space shuttle. To find the mass, we use Newton's second law of motion, which states that Force = [tex]mass imes acceleration (F = m imes a).[/tex] However, we first need to convert the given acceleration from miles per hour squared to meters per second squared, and then convert the mass obtained in kilograms to pounds-mass.

First, let's convert the acceleration: 18,000 miles/hour2 is approximately 8,046.72 m/s2 (using the conversion factor 1 mile = 1,609.34 meters and 1 hour = 3600 seconds). Next, we can calculate the mass (in kilograms) by rearranging the formula to m = F / a, which gives us 53,000 N / 8,046.72 m/s2
= approximately 6.59 kilograms. We then convert kilograms to pounds-mass (1 kilogram = 2.20462 pounds), resulting in the mass of the spacecraft being approximately 14.52 pounds-mass.

The mass of the spacecraft is approximately [tex]261,313.24\ pounds[/tex] mass.

To find the mass of the spacecraft, we can use Newton's second law of motion:

[tex]\[ F = ma \][/tex]

where:

[tex]- \( F \)[/tex] is the force (thrust) provided by the first stage ([tex]53 kN[/tex]),

[tex]- \( m \)[/tex] is the mass of the spacecraft,

[tex]- \( a \)[/tex] is the acceleration ([tex]18,000 \ mph^2[/tex]).

First, let's convert the thrust from kilo-newtons (kN) to newtons (N), since the unit of acceleration is meters per second squared (m/s²):

[tex]\[ 1 \text{ kN} = 1000 \text{ N} \][/tex]

So, [tex]53 kN[/tex] is equivalent to [tex]\(53 \times 1000 = 53000\) N.[/tex]

Now, let's convert the acceleration from miles per hour squared (mph²) to meters per second squared (m/s²).

[tex]\[ 1 \text{ mph} = \frac{1609.34}{3600} \text{ m/s} \][/tex]

[tex]1 mph^2 = \left(\frac{1609.34}{3600}\right)^2 \text{ m/s²}[/tex]

[tex]1 mph^2 = 0.447 m/s^2}[/tex]

We can rearrange Newton's second law to solve for the mass [tex]\( m \)[/tex]

[tex]\[ m = \frac{F}{a} \][/tex]

[tex]\[ m = \frac{53000 \text{ N}}{0.447 \text{ m/s²}} \][/tex]

[tex]\[ m = 118488 \text{ kg} \][/tex]

To convert kilograms to pounds-mass, we use the conversion factor:

[tex]\[ 1 \text{ kg} = 2.20462 \text{ pounds-mass} \][/tex]

So,

[tex]\[ \text{mass (pounds-mass)} = 118488 \text{ kg} \times 2.20462 \][/tex]

[tex]\[ \text{mass (pounds-mass)} = 261313.24 \text{ pounds-mass} \][/tex]

The electric field strength between parallel conducting plates can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. For a spacing of 1.00 mm, the electric field strength is 7.08 × 10^-8 N/C. For a spacing of 2.00 mm, the electric field strength is 3.54 × 10^-8 N/C.

The electric field strength between parallel conducting plates can be calculated using the formula:

E = V/d

Where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. In this case, the potential difference is given as 0.708 nC, or 0.708 × 10-9 C.

To find the electric field strength, we need to convert the charge to coulombs and divide it by the spacing between the plates. For a spacing of 1.00 mm (or 0.01 cm), the electric field strength is:

E = (0.708 × 10-9 C) / (0.01 cm) = 7.08 × 10-8 N/C

For a spacing of 2.00 mm (or 0.02 cm), the electric field strength is:

E = (0.708 × 10-9 C) / (0.02 cm) = 3.54 × 10-8 N/C

The electric field strength inside the capacitor is 2000 N/C. The potential difference is 2 V for a 1.00 mm separation and 4 V for a 2.00 mm separation between the plates.

To find the electric field strength (E) and potential difference (V) across a parallel-plate capacitor, we use the following steps:

Therefore, the electric field strength inside the capacitor is 2000 N/C, the potential difference is 2 V for 1.00 mm separation, and 4 V for 2.00 mm separation.

Answer:

238 Degree

Explanation:

Data given

F1=(3.3,-0.5) and F2=(-3.8,-0.3).

To determine the angle F1+F2 makes with the positive x-axis, we need to determine the magnitude of the force F1+F2.

Since force is a vector quantity, we add the vectors component by component

[tex]F_{1}+F_{2}=<3.3,-0.5> +<-3.8,-0.3>\\F_{1}+F_{2}=<3.3+(-3.8), -0.5+(-0.3)>\\ F_{1}+F_{2}=<-0.5,-0.8>\\F_{1}+F_{2}=(-0.5,-0.8)[/tex]

To determine the angle, we use

[tex]F=(x,y)\\\alpha=arctan(\frac{y}{x} )\\Hence for \\F_{1}+F_{2}=(-0.5,-0.8)\\\alpha=arctan(\frac{-0.8}{-0.5} )\\\alpha=58^{0}[/tex]

Since the component of the force F1+F2 is a negative y and negative x which are located in the  3rd quadrant,  the angle can be calculated as

∝=58+180=238 degree

Hence The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees is 238 Degree

The angle measured counterclockwise from the positive x-axis is θ = 57.99°

Here we know that:

First, we need to add the forces, we will get:

F1 + F2 =  (3.3, -0.5) + (-3.8, -0.3) = (3.3 - 3.8, -0.5 - 0.3))

F1 + F2 = (-0.5, -0.8)

Now, the angle measured counterclockwise from the positive x-axis of a vector

(a, b) is given by:

θ = Atan(b/a).

Where Atan(x) is the inverse tangent function.

So in this case the angle will be:

θ = Atan(-0.8/-0.5) = 57.99°

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Answer:

0.12959085 J

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

[tex]U=k\dfrac{q^2}{d}[/tex]

In this system with three charges which are equidistant from each other

[tex]U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}[/tex]

[tex]\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J[/tex]

The potential energy of the system is 0.12959085 J

Answer:

Explanation:

We are given that Sarah throws a tennis ball as far as she can.

At the moment the ball reaches its maximum height.

We have to find the true statement if air resistance is neglect.

When air resistance is negligible then the force act on the ball is force due to gravity.

The ball throw vertically then the acceleration act on the ball is acceleration due to gravity.

The value of g=-9.8 m/square sec

It acts on the ball in downward direction .

Therefore, the ball's vertical acceleration is downwards.

The horizontal acceleration is zero because the ball reaches at maximum height then there is no force which act in horizontal direction on the ball.

Therefore, horizontal acceleration of the ball is zero.

Hence, option a and e are true.

The statement B is correct which states that "The ball's vertical acceleration is downwards".

The statement E is correct which states that "The ball's horizontal acceleration is zero".

The given condition is that Sarah throws a tennis ball as far as she can. Also given that the air resistance is negligible.

Let consider that the ball is thrown vertically. In this case, the force on the ball is the force due to gravitational acceleration on the ball.

Hence the vertical acceleration on the ball is equivalent to the gravitational acceleration of 9.8 m/s2. The vertical force acting on the ball will be in a downward direction because this is the force due to gravity.

Hence the statement B is correct which states that "The ball's vertical acceleration is downwards".

When the ball reaches its maximum height, the horizontal acceleration will be zero on the ball, as there is no force acting on the ball in the horizontal direction.

Hence the statement E is correct which states that "The ball's horizontal acceleration is zero".

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Answer:

The correct option is that both the ends will remain neutral.

Explanation:

As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.

The correct option is: There is negative charge spread evenly on both ends.

 When a conducting rod is brought into contact with a charged ball, charge transfer occurs due to the process of induction and conduction. Initially, when the charged ball (let's assume it is positively charged) is brought close to the rod, it will induce a negative charge on the end of the rod nearest to the ball (end A) and a positive charge on the far end (end B) due to the separation of charges within the conductor. This is because the positive charge on the ball will repel the positive charges within the rod, pushing them away, and attract the negative charges, pulling them towards the ball.

 Once the rod is actually touched to the ball, the negative charges on the rod (which were induced by the ball's positive charge) will be neutralized by the positive charges from the ball, leaving the rod with a net negative charge after the ball is moved away. This negative charge will be evenly distributed along the rod initially.

 However, since the rod is not grounded, the negative charge will not have a path to escape and will remain on the rod. Due to the repulsion between like charges (negative-negative), these charges will repel each other and spread out as far as possible from each other along the length of the rod. This means that the negative charge will be spread evenly on both ends of the rod, as this arrangement minimizes the repulsive forces between the charges and results in the lowest potential energy configuration.

 Therefore, the final arrangement of any charge on the rod, when the charged ball is far away, will be an even distribution of negative charge on both ends A and B of the rod. This is consistent with the principle of electrostatic equilibrium, where charges within a conductor will redistribute themselves to minimize repulsion and achieve the lowest energy state.

We have that the  the rock's initial launch speed  is mathematically given as

From the question we are told

Generally the equation for the Motion   is mathematically given as

[tex]H=ut+0.5gt^2\\\\Therefore\\\\4.9=0*t+0.5*9.8*t^2\\\\t=1s\\\\Therefore\\\\sx=uxt\\\\20=ux*1\\\\[/tex]

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To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, and it travels 20 m horizontally in 1 second, the initial launch speed can be calculated as 20 m/s.

To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, its initial vertical velocity is zero. The only force acting on the rock in the vertical direction is gravity, which causes it to accelerate downward at a rate of 9.8 m/s². Using the formula:

-yf = yi + vit - 1/2gt²

where yi represents the initial height and yf represents the final height, we can solve for the initial downward velocity. Since the rock reaches the ground (yf = 0) and starts at a height of 4.9 m, the equation becomes:

0 = 4.9 + 0 - 1/2(9.8)t²

Simplifying the equation gives:

t² = 4.9/4.9

t = 1 second.

Since the horizontal distance traveled by the rock is 20 m, and it takes 1 second to reach the ground, the horizontal speed (initial launch speed) of the rock can be calculated using the formula:

v = d/t = 20 m/1 s = 20 m/s.

Therefore, the rock's initial launch speed is 20 m/s.

Answer:

The electric flux through each of the 3 sides is 2.95 Wb

Solution:

As per the question:

Length of the edges, l = 27.0 cm = 0.27 m

Strength of the electric field, E = 280 N/C

To calculate the electric field through each of the three sides:

Area of the equilateral triangle is given by:

[tex]A = \frac{\sqrt{3}}{4}l^{2}[/tex]

[tex]A = \frac{\sqrt{3}}{4}\times (0.27)^{2} = 0.0316\ m^{2}[/tex]

Now, the electric flux that passes through the base is given by:

[tex]\phi = - E\cdot A = - EA[/tex]

[tex]\phi = 280\times 0.0316 = - 8.85\ Wb[/tex]

Now, the overall flux that passes through the surface and the base of the tetrahedron is zero.

[tex]\phi_{total} = \phi + 3phi_{surface}[/tex]

[tex]0 = \phi + 3phi_{surface}[/tex]

[tex]phi_{surface} = -\frac{\phi }{3}[/tex]

[tex]phi_{surface} = -\frac{- 8.85}{3} = 2.95\ Wb[/tex]

Final answer:

A 26-N force must act down and parallel to the incline for the bin to move down at constant velocity since this force would balance out the 26-N frictional force acting up the incline.

Explanation:

The subject of this question is Physics, specifically the concepts related to mechanics and forces on inclines with friction. The student is in High School level, most likely studying the fundamentals of Newtonian mechanics.

Given that the bin is already being pushed up the incline with a 26-N force and moves at a constant speed, this means the net force on the bin is zero, therefore the force of friction must also be 26 N but acting downwards along the incline. When the bin moves down at a constant velocity, the force of friction still acts up the incline (opposite to the bin's movement) and therefore, to maintain a constant velocity, the force applied must be equal in magnitude to the frictional force but directed down the incline. Hence, a 26-N force must be applied down and parallel to the incline for the bin to move down the incline at constant velocity.

Answer:

29.79 N/m

Explanation:

A spring connected in series behaves like a capacitor connected in series.

Note: Spring and capacitor are alike because they both store energy, While the former store mechanical energy, the later store electrical energy.

From the above, the effective spring connected in series is given by the formula below

1/Kt = 1/K1 + 1/K2 ........................ Equation 1

Where Kt = effective spring constant, K1 = spring constant of spring 1, K2 = spring constant of spring 2

Making K2 the subject of the equation,

K2 = KtK1/(K1-Kt)..................... Equation 2

Given: K1 = 61 N/m, Kt = 20 N/m.

Substitute into equation 2

K2 = (61×20)/(61-20)

K2 = 1220/41

K2 = 29.76 N/m.

Hence the spring constant in the second spring = 29.79 N/m

The spring constant for Spring #2 is 369.17 N/m.

To find the spring constant of Spring #2, we need to use the formula for the effective spring constant of springs in series. The formula is:

1/keff = 1/k1 + 1/k2

Given that k1 = 61.0 N/m and the effective spring constant (keff) is 20.0 N/m, we can plug these values into the formula and solve for k2:

1/20 = 1/61 + 1/k2

Now we can solve for k2:

1/k2 = 1/20 - 1/61

k2 = 1/(1/20 - 1/61)

k2 = 369.17 N/m

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Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

Answer

given,

Volume of Depression, V = 75000 m³

borrow soil

Density of soil,γ = 1540 Kg/m³

water content,w = 8 % = 0.08

Specific gravity of solid,G = 2.66

Final in-place

dry density = 1790 kg/m³

water content = 13 % = 0.13

a) Volume of borrow soil require to fill the depression

   Mass of solid solid,m= dry density of inplace soil x Volume of depression

                              = 1790 x 75000

                           m = 1.3425 x 10⁸ Kg

dry density of the borrow pit

 [tex]\gamma_d = \dfrac{\gamma}{1 + w}[/tex]

 [tex]\gamma_d = \dfrac{1540}{1 + 0.08}[/tex]

 [tex]\gamma_d = 1425.93\ kg/m^3[/tex]

Volume of borrow soil required

    [tex]V = \dfrac{m}{\gamma_d}[/tex]

    [tex]V = \dfrac{1.3425\times 10^8}{1425.93}[/tex]

           V = 94149 m³

b) Water required

  [tex]W = 1790\times 75000\times 0.13 - 1425.93\times 94149\times 0.08[/tex]

 W = 6.71 x 10⁶ Kg

Water required to achieve 13% moisture is equal to W = 6.71 x 10⁶ Kg

Final answer:

To fill a 75,000 m³ depression, 93,590.46 m³ of borrow soil is required when accounting for the dry density change. To achieve a final water content of 13%, an additional 17,452,500 kg of water mass is needed.

Explanation:

To answer this question, we must first understand the relationship between the initial and final states of the borrow soil in terms of their volumes, densities, and moisture contents. The question is divided into two parts: calculating the volume of borrow soil needed and determining the water mass addition required to achieve the desired moisture content.

a) How many m³ of borrow soil are needed to fill the depression?

Since we want to fill a depression with a volume of 75,000 m³, we need to consider the change in density from the borrow state to the in-place dry density. The in-place dry density is given as 1790 kg/m³. The dry density of the borrow soil can be calculated by subtracting the mass of water from the total mass.

First, we convert the water content to a decimal by dividing by 100:
Water content = 8% / 100 = 0.08

Using the formula for dry density (
dry density = total density / (1 + water content)), we find:
Dry density of borrow soil = 1540 kg/m³ / (1 + 0.08) = 1425.93 kg/m³

The volume of the borrow soil needed can be calculated by the volume of the depression divided by the ratio of the final in-place dry density to the dry density of the borrow soil:
Volume of borrow soil needed = 75,000 m³ x (1790 kg/m³ / 1425.93 kg/m³) = 93,590.46 m³

b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?

For the final water content of 13%, we'll use the in-place dry density to determine the total mass, then calculate the water mass needed:
Total in-place mass = volume x in-place dry density
Total in-place mass of fill = 75,000 m³ x 1790 kg/m³= 134,250,000 kg

Convert 13% to decimal:
Water content = 13% / 100 = 0.13

We want 13% of the total mass to be water, so the water mass required is:
Water mass = total mass x water content
Water mass needed = 134,250,000 kg x 0.13 = 17,452,500 kg

This is the additional water mass required to achieve the desired moisture content of 13%.

Answer with Step-by -step explanation:

We are given that

b.[tex]\mid A\mid=46 m[/tex]

[tex]\theta=20^{\circ}[/tex] below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=[tex]x=360-20=340^{\circ}[/tex]

x-component of vector A=[tex]A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24[/tex]

y-Component of vector A=[tex]A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64[/tex]

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis=[tex]x'=42^{\circ}[/tex]

x-component of vector B=[tex]B_x=86cos42=63.64[/tex]

y-Component of vector B=[tex]B_y=86sin42=57.62[/tex]

Vector A=[tex]A_xi+A_yj=43.24i-15.64j[/tex]

Vector B=[tex]B_xi+B_yj=63.64i+57.62j[/tex]

Vector C=A+B

Substitute the values

[tex]C=43.24i-15.64j+63.64i+57.62j[/tex]

[tex]C=106.88i+41.98j[/tex]

c.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}[/tex]

The direction of the vector C=21.5 degree

To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, start by drawing the x and y axes. Use the component method of vector addition to find the magnitude of C⃗. Use the inverse tangent function to find the direction of C⃗.

To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, we start by drawing the x and y axes. Vector A⃗ has a magnitude of 46.0 m and points 20.0° below the positive x-axis, so we draw A⃗ starting from the origin and make an angle of 20.0° with the positive x-axis in a downward direction. Vector B⃗ has a magnitude of 86.0 m and points 42.0° above the negative x-axis, so we draw B⃗ starting from the origin and make an angle of 42.0° with the negative x-axis in an upward direction. To find the vector C⃗=A⃗+B⃗, we add the x-components and the y-components of A⃗ and B⃗ separately. Then we use the Pythagorean theorem to find the magnitude of C⃗ and the inverse tangent function to find the direction of C⃗ in relation to the positive x-axis.

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Answer:

assume nitrogen is an ideal gas with cv=5R/2

assume argon is an ideal gas with cv=3R/2

n1=4moles

n2=2.5 moles

t1=75°C   in kelvin t1=75+273

t1=348K

T2=130°C  in kelvin t2=130+273

t2=403K

u=пCVΔT

U(N₂)+U(Argon)=0

putting values:

=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)

by simplifying:

Tfinal=363K

To determine the change in entropy when the gases mix adiabatically and completely, we'll follow these steps:

1. Calculate the initial entropy of each gas before mixing.

2. Calculate the final entropy of the combined gases after mixing.

3. Calculate the change in entropy.

Given information:

For nitrogen gas:

- Moles of nitrogen gas, [tex]n_N2[/tex] = 4 mol

- Temperature of nitrogen gas, [tex]T_N2[/tex] = 75°C = 75 + 273.15 K

- Pressure of nitrogen gas, [tex]P_N2[/tex] = 30 bar = 30 * 100 kPa

For argon gas:

- Moles of argon gas, [tex]n_Ar[/tex] = 2.5 mol

- Temperature of argon gas, [tex]T_Ar[/tex] = 130°C = 130 + 273.15 K

- Pressure of argon gas, [tex]P_VR[/tex] = 20 bar = 20 * 100 kPa

Given specific heat capacities :

- [tex]CV_N2[/tex] = (5/2)R for nitrogen gas

- [tex]CV_Ar[/tex] = (3/2)R for argon gas

The change in entropy, ΔS, can be calculated using the equation:

[tex]\[ΔS = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_f}{T_{\text{N2}}}\right) + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_f}{T_{\text{Ar}}}\right)\][/tex]

where [tex]\(T_f\)[/tex]  is the final temperature after mixing.

1. Calculate the initial entropy of each gas:

[tex]\[S_{\text{N2}} = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_{\text{N2}}}{T_0}\right)\][/tex]

[tex]\[S_{\text{Ar}} = n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_{\text{Ar}}}{T_0}\right)\][/tex]

where [tex]\(T_0\)[/tex] is the reference temperature (usually taken as 298.15 K).

2. Calculate the final temperature after mixing using the adiabatic process equation:

[tex]\[T_f = \left(\frac{n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot T_{\text{N2}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot T_{\text{Ar}}}{n_{\text{N2}} \cdot C_{V_{\text{N2}}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}}}\right)\][/tex]

3. Calculate the change in entropy using the formula above.

Let's plug in the values and calculate the change in entropy. First, we'll calculate the initial entropies of each gas:

[tex]\[S_{\text{N2}} = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{75 + 273.15}{298.15}\right)\][/tex]

[tex]\[S_{\text{Ar}} = 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{130 + 273.15}{298.15}\right)\][/tex]

Now, calculate [tex]\(T_f\)[/tex] using the adiabatic process equation:

[tex]\[T_f = \left(\frac{4 \times \left(\frac{5}{2}\right)R \times 75 + 2.5 \times \left(\frac{3}{2}\right)R \times 130}{4 \times \left(\frac{5}{2}\right)R + 2.5 \times \left(\frac{3}{2}\right)R}\right)\][/tex]

Finally, calculate the change in entropy:

[tex]\[ΔS = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{T_f}{75 + 273.15}\right) + 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{T_f}{130 + 273.15}\right)\][/tex]

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Explanation:

As the force is given as 5.30 kN or [tex]5.30 \times 1000 N[/tex]. Hence, mass will be calculated as follows.

           [tex]F_{w}[/tex] = mg

         [tex]5.30 \times 10^{3} = m \times 9.8 m/s^{2}[/tex]

                   m = 540.816 kg

(a)  At the top, centripetal force [tex]F_{c}[/tex] is acting upwards and the weight of the riders and car, [tex]F_{w}[/tex] will be acting downwards.

Therefore, force on the car by the boom will be calculated as follows.

                   [tex]F_{B} = F_{w} - F_{c}[/tex]

or,          [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]

                         = [tex]5.30 \times 1000 N - \frac{540.816 kg \times (3.60)^{2}}{12}[/tex]

                         = 4715.919 N

Hence, the force [tex]F_{B}[/tex] on the car from the boom is 4715.919 N. This means that the car will be hanging on the boom and the boom will exert an upward force.

(b)   Now at the top, centripetal force [tex]F_{c}[/tex] will be acting upwards and the weight of cars and car riders will be acting in the downwards direction.

Hence, we will calculate the force on car by the boom as follows.

       [tex]F_{B} = F_{w} - F_{c}[/tex]

or,         [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]

                        = [tex]5.30 \times 1000 N - \frac{540.816 kg \times (14.0)^{2}}{12}[/tex]

                        = -3533.33 N

Therefore, car will be pushing on the boom and the boom will exert a downward force.  

Answer:

The constant force exerted on the bullet is 1590.87 N.

Explanation:

It is given that,

Mass of the bullet, m = 8.6 g

Initial speed of the bullet, u = 0          

Final speed of the bullet, v = 430.1 m/s

We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,

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

[tex]v^2=2ad[/tex]

[tex]a=\dfrac{v^2}{2d}[/tex]

[tex]a=\dfrac{(430.1)^2}{2\times 0.5}[/tex]

[tex]a=184986.01\ m/s^2[/tex]

Let F is the force exerted. It is given by :

[tex]F=ma[/tex]

[tex]F=8.6\times 10^{-3}\times 184986.01[/tex]

F = 1590.87 N

So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.                                                  

Answer:

435.467980296 Hz

Explanation:

T = Temperature = 37.0ºC

L = Length of the tube = 0.203 m

Speed of sound at a specific temperature is given by

[tex]v=331.4+0.6\times T\\\Rightarrow v=331.4+0.6\times 37\\\Rightarrow v=353.6\ m/s[/tex]

Frequency is given by (one end open other end closed)

[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{353.6}{4\times 0.203}\\\Rightarrow f=435.467980296\ Hz[/tex]

The fundamental frequency is 435.467980296 Hz

The fundamental frequency of a tube, considered as an approximation for the vocal apparatus and closed at one end, with a length of 0.203 meters and at an air temperature of 37°C, is calculated using the formula for frequency f=Vw/(4L). Substituting in the values, the fundamental frequency is about 432 Hz.

The given question involves a concept from physics, specifically sound waves and resonance. When considering the breathing passages and mouth as a resonating tube closed at one end, the fundamental resonant frequency can be calculated using the formula Vw = fa, where Vw is the speed of sound, f represents frequency, and a is the wavelength.

For a tube closed at one end, the wavelength (λ) is four times the length of the tube. So, λ = 4L. We can rearrange the formula to solve for fundamental frequency: f = Vw / a = Vw / (4L).

If the air temperature is 37°C, the speed of sound (Vw) in air is approximately 351 m/s. Substituting these values in, you get: f = 351m/s / (4*0.203m) ≈ 432 Hz. So, the fundamental frequency of this tube (our vocal apparatus approximation) at 37°C would be about 432 Hz.

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Answer: (D) Her weight is less, and her mass is the same.

Explanation:

Mass is the total amount of matter contained in the body.

Weight is the force exerted by gravity on a mass.

[tex]weight=mass\times gravity[/tex]

When the value of g is more, weight is more.

For moon g= [tex]1.6m/s^2[/tex] , whereas for earth [tex]g=9.8m/s^2[/tex]

Thus weight is more on earth as compared to moon.

On the moon, compared to at home on earth: Her weight is less, and her mass is the same.

Answer:

Explanation:

Weight is defined as the force with which a planet pulls the object towards its centre.

Weight = mass  x acceleration due to gravity of the planet

W = m x g

The amount of matter contained in the substance is called mass.

the mass of the substance remains same at every planet but the weight is changed as the value of acceleration due to gravity is different for all the planets.

As the acceleration due to gravity on moon is 1.6 m/s^2 and it is less than the acceleration due to gravity on earth. So, the weight of the astronaut is less than the weight on earth and mass remains same.

Weight is less but the mass is same.

Option (D) is true.

Answer:

v_th = 3.1 * 10^5 m/s

Explanation:

Given:

- mass of 238-Uranium m_u = 3.952 *10^-25 kg

- mass of alpha particle m_a = 6.64 * 10^-27 kg

- mass of thorium particle m_th = 3.885*10^-25 kg

- velocity of 238-Uranium v_u = 5.0 *10^5 m/s

- velocity of alpha particle v_a = 1.4 *10^7 m/s

Find:

- The recoil velocity of the thorium particle.

Solution:

- To solve this problem we will use conservation of momentum in both x and y direction.

- Momentum conservation in x-direction:

                                        P_i = P_f

                  m_u*v_u = m_a*v_a*cos(Q) + m_th*v_th,x

where v_th,x is the x component of thorium velocity:

                 P_i = 3.952 *10^-25 * 5.0 *10^5 = 1.976*10^-19

      P_f = 6.64 * 10^-27*1.4 *10^7*cos(25.4) + 3.885*10^-25*v_th,x

                 P_f = 8.3974*10^-20 + 3.885*10^-25*v_th,x

Hence,

             1.976*10^-19 = 9.296*10^-20 + 3.885*10^-25*v_th,x

                         v_th,x = 2.92473 * 10^5 m/s        

- Momentum conservation in y-direction:

                                        P_i = P_f

                         0 = m_a*v_a*sin(Q) + m_th*v_th,y

where v_th,x is the x component of thorium velocity:

                         v_th,y = m_a*v_a*sin(Q) / m_th    

        v_th,y = 6.64 * 10^-27*1.4 *10^7*sin(25.4) / 3.885*10^-25

Hence,

                            v_th,y = 1.02635 * 10^5 m/s        

- The magnitude of recoil velocity:

                        v_th = sqrt ( v_th,x ^2 + v_th,y ^2 )

            v_th = sqrt ( (2.92473 * 10^5)^2 + (1.02635 * 10^5)^2 )

                                  v_th = 3.1 * 10^5 m/s

Answer:

[tex]A'=2A[/tex]

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

[tex]E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}[/tex]

When the spring is in its equilibrium position, that is [tex]x=0[/tex], the object speed its maximum. So, we have:

[tex]\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}[/tex]

In order to double its maximum speed, that is [tex]v'{max}=2v_{max}[/tex]. We have:

[tex]A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A[/tex]

The factor we must increase the amplitude of the vibration to double its maximum speed is 2.

Apply the principle of conservation of energy to determine the amplitude;

[tex]U = K.E\\\\\frac{1}{2} KA^2 = \frac{1}{2}mv^2\\\\KA^2 = mv^2\\\\A^2 = \frac{m}{K} v^2\\\\\frac{A_1^2}{v_1^2 } = \frac{A_2^2}{v_2^2 } \\\\A_2 = \frac{v_2 A_1}{v_1} \\\\A_2 = \frac{2v_1 \times A_1}{v_1} \\\\A_2 =2A_1[/tex]

Thus, the factor we must increase the amplitude of the vibration to double its maximum speed is 2.

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Answer:

Explanation:

=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!

Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.

Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From

Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?

a.There is no energy stored in the string: The string will remain straight for all subsequent times.

b.Energy will flow into the string, causing the standing wave to form at a later time.

c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.

d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.

Here, the spatial and temporal parts of the wave function are A sin(kx) and sin(ωt), respectively. The displacement of the string at x=0 is 0. Although the string is straight at time t=π/2ω, there is still energy stored in it due to parts of the string having nonzero velocity.

Part C: The spatial part of the wave function, ye(x), can be expressed as A sin(kx), and the temporal part, yt(t), is represented by sin(ωt). These expressions are valid for a wave of unit amplitude.

Part E: At position x=0, the displacement of the string is 0. This occurs because, for the wave equation, when inputting x=0 into the spatial part, the value for y becomes 0.

Part G: The correct answer is option c. Although the string is straight at time t=π/2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string. Option d is incorrect as the total mechanical energy in a perfectly straight string does not oscillate, but remains constant.

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