An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

[tex]\boxed{y=k\frac{pd}{w}}[/tex]

Let's explain what direct and indirect variation mean:

[tex]y=kxw[/tex] for some nonzero constant [tex]k[/tex] that is the constant of variation or the constant of proportionality.

[tex]y=\frac{k}{x}[/tex] for some nonzero constant [tex]k[/tex], where [tex]k[/tex] is also the constant of variation.

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In this problem, [tex]u[/tex] varies jointly with [tex]p[/tex] and [tex]d[/tex] and inversely with [tex]w[/tex], being [tex]k[/tex] the constant of proportionality, then:

[tex]\boxed{y=k\frac{pd}{w}}[/tex]

The equation expressing the described relationship is u = kpd/w. This represented u varying jointly with p and d and inversely with w, with k being the constant of proportionality.

The equation that expresses the relationship where 'u' varies jointly with 'p' and 'd' and inversely with 'w' using 'k' as the constant of proportionality would be: u = kpd/w. Here, 'k' is the constant of proportionality which determines the rate at which 'u' varies relative to 'p', 'd', and 'w'. Joint variability is expressed by multiplying the variables 'p' and 'd', while inverse proportionality is shown through dividing by 'w'.

This equation effectively expresses the described relationship - as 'p' or 'd' increase (or 'w' decrease), 'u' increases and vice versa. It’s important to understand that the constant 'k' would need to be determined based on additional context or data.

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Answer:

[tex]5.5\cdot 10^{-11} C[/tex]

Explanation:

The capacitance of the parallel-plate capacitor is given by:

[tex]C=\epsilon_0 \frac{A}{d}[/tex]

where

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

[tex]A=190 mm^2 = 190 \cdot 10^{-6} m^2[/tex] is the area of the plates

[tex]d=1.2 mm = 0.0012 m[/tex] is the separation between the plates

Substituting,

[tex]C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F[/tex]

The energy stored in the capacitor is given by

[tex]U=\frac{Q^2}{2C}[/tex]

Since we know the energy

[tex]U=1.1 nJ = 1.1 \cdot 10^{-9} J[/tex]

we can re-arrange the formula to find the charge, Q:

[tex]Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C[/tex]

The charge needed to be transferred between the plates to store 1.1 nJ of energy is approximately 55.5 pC.

To determine the charge that must be transferred between two initially uncharged parallel plates to store 1.1 nJ of energy, we use the concept of a parallel plate capacitor. The energy stored in a capacitor is given by the formula:

U = 0.5 * C * V²

Where U is the energy (1.1 nJ = 1.1 × 10⁻⁹J), C is the capacitance, and V is the potential difference between the plates.

Step 1: Calculate the Capacitance

Capacitance (C) for a parallel-plate capacitor is given by:

C = (0 * A) / d

Where 0 is the permittivity of free space (8.85 × 10⁻¹² C²/N·m²), A is the area of the plates (190 mm² = 190 × 10⁻⁶ m²), and d is the separation between the plates (1.2 mm = 1.2 × 10⁻³ m).

Substitute the values:

C = (8.85 × 10⁻¹²* 190 × 10⁻⁶) / 1.2 × 10⁻³

C ≈ 1.4 × 10⁻¹² F

Step 2: Calculate the Voltage

Using the energy formula, solve for V:

1.1 × 10⁻⁹ J = 0.5 * 1.4 × 10⁻¹² F * V²

V² = (1.1 × 10⁻⁹ J) / (0.5 * 1.4 × 10⁻¹²F)

V² ≈ 1571 J/F

V ≈ 39.65 V

Step 3: Calculate the Charge

Use the relationship Q = C * V:

Q = 1.4 × 10⁻¹² F * 39.65 V

Q ≈ 5.55 × 10⁻¹¹ C

Therefore, approximately 55.5 pC (picoCoulombs) of charge must be transferred from one plate to the other.

To calculate the time required to deposit 0.500 g of metallic nickel on a custom car part using a current of 3.00 A, you can use the equation Time (s) = Mass (g) / (Current (A) × Charge (C/g)). Calculations involving the charge of nickel and the given values can be used to determine the time in seconds. Converting the time to minutes or hours is also possible.

To calculate the time required to deposit 0.500 g of metallic nickel using a current of 3.00 A, we need to use the equation:

Time (s) = Mass (g) / (Current (A) × Charge (C/g))

For nickel, the charge is 2+ and the atomic mass is approximately 58.69 g/mol. From these values, we can calculate the charge in C/g:

Charge (C/g) = (2 × 1.602 × 10-19 C) / (58.69 g/mol)

Using this value and the given mass and current, we can calculate the time required in seconds:

Time (s) = 0.500 g / (3.00 A × Charge (C/g))

Converting the time to minutes or hours can be done by dividing by 60 or 3600, respectively.

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Answer:

145.8 m

Explanation:

M = mass of the runaway train car = 15,000 kg

V₀ = initial speed of the train = 5.4 m/s

V = final speed of the train = 0 m/s

F = force applied to bring the car to rest = 1500 N

d = distance traveled before stopping

Using work-change in kinetic energy theorem

Work done by applied force = Change in kinetic energy

- F d = (0.5) M (V² - V₀²)

- (1500) d = (0.5) (15000) (0² - 5.4²)

d = 145.8 m

We apply Newton's second law to find the deceleration of the train car. Then, we use the work-energy principle to compute the distance traveled during the stopping process.

The problem at hand involves using concepts of physics, specifically Newton's second law and work-energy principle. The mass of the train car is 15000 kg and its initial speed is 5.4 m/s. The external force acting on it is 1500 N which brings the car to rest, implying that its final speed is 0m/s.

To solve the problem, we first need to find the deceleration of the train car using the formula F=ma (where F=force, m=mass, a=acceleration). The negative sign indicates deceleration. After calculating the deceleration, we will use the equation of motion to find the distance traveled during the stopping period. The equation is v2 = u2 + 2as where v is final speed, u is initial speed, a is acceleration and s is distance.

After substituting the known values into the equation, we will be able to solve for s (the distance covered while stopping).

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Answer:

Emf through a single coil of wire is 588 V.

Explanation:

We need to find the emf through a single coil of wire if the magnetic flux changes from -57 Wb to +43 Wb in 0.17 s

According to Faraday's law, emf of coil is directly proportional to the rate of change of magnetic flux. It can be written as :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

Initial flux, [tex]\phi_1=-57\ Wb[/tex]

Final flux, [tex]\phi_2=+43\ Wb[/tex]

So, [tex]\epsilon=\dfrac{\phi_2-\phi_1}{dt}[/tex]

[tex]\epsilon=\dfrac{43\ Wb-(-57\ Wb)}{0.17\ s}[/tex]

[tex]\epsilon=588.23\ V[/tex]

or

[tex]\epsilon=588\ V[/tex]

So, the EMF through a single coil of wire is 588 V. Hence, this is the required solution.

Answer:

208 N

Explanation:

The net force on the woman is equal to the centripetal force, which is given by

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

where

m is the mass of the woman

v is her speed

r is the radius of the wheel

here we have:

r = d/2 = 5 m is the radius of the wheel

[tex]m=\frac{W}{g}=\frac{670 N}{9.8 m/s^2}=68.4 kg[/tex] is the mass of the woman (equal to her weight divided by the acceleration of gravity)

The wheel makes one revolution in t=8.0 s, so the speed is:

[tex]v=\frac{2\pi r}{t}=\frac{2\pi (5.0 m)}{8.0 s}=3.9 m/s[/tex]

so now we can find the centripetal force:

[tex]F=(68.4 kg)\frac{(3.9 m/s)^2}{5.0 m}=208 N[/tex]

Answer:

5 units at 203 degrees

Explanation:

The equilibrant vector must have components that are opposite to those of the initial vector.

The components of the initial vector are:

[tex]v_x = v cos \theta = 5 cos 23^{\circ}=4.60\\v_y = v sin \theta = 5 sin 23^{\circ} = 1.95[/tex]

So the components of the equilibrant vector must be

[tex]v_x = -4.60\\v_y = -1.95[/tex]

which means its magnitude is

[tex]v=\sqrt{v_x^2 + v_y^2}= 5[/tex] (same magnitude as the initial vector)

and it is located in the 3rd quadrant, so its angle will be

[tex]\theta = 180^{\circ} + tan^{-1} (\frac{1.95}{4.60})=203^{\circ}[/tex]

Answer: a. 1.203 m/s   b.0.35m

Explanation:

The velocity [tex]V[/tex] of a wave is given by the following equation:

[tex]V=\frac{\lambda}{T}[/tex]  (1)

Where:

[tex]\lambda=6.50m[/tex] is the wavelength (the horizontal space between crest and crest)

[tex]T=2.70s.2=5.4s[/tex] is the period of the wave (If we were told it takes a time of 2.70 s for the boat to travel from its highest point to its lowest, the period is twice this amount, which is the time it takes to travel one wavelength)

Substituting the known values:

[tex]V=\frac{6.50m}{5.4s}=1.203m/s[/tex]  (2)

[tex]V=1.203m/s[/tex]  Velocity of the wave

The amplitude [tex]A[/tex] of a wave is defined is given by the following equation:

[tex]A=\frac{maximumpoint-minimumpoint}{2}[/tex]  (3)

If we know the total distance between the highest point to the lowest point is 0.7 m. This means:

[tex]maximumpoint-minimumpoint=0.7m[/tex]

Substituting this value in (3):

[tex]A=\frac{0.7m}{2}[/tex]  (4)

[tex]A=0.35m[/tex]  This is the amplitude of the wave

The waves for the boat of fisherman is travelling with the speed of 1.203 m/s and the amplitude of each wave is 0.35 m.

Speed of wave is the rate of speed by which the wave travel the distance in the time taken by it.

The distance traveled by the boat  from its highest point to the lowest point is 0.700 m. The time taken by the boat is 2.70.

The fisherman sees that the wave crests are spaced a horizontal distance of 6.50 m apart. As, the horizontal distance of crests is the wavelength of the wave. Thus, the wavelength of the wave is,

[tex]\lambda=6.50\rm s[/tex]

The period of the wave is twice the time taken from highest to lowest point. Thus, the time period of the wave is,

[tex]T=2\times2.7\\T=5.4\rm s[/tex]

The speed of the wave is the ratio of wavelength to the time period. Thus, the speed of the waves travelling is,

[tex]v=\dfrac{6.50}{5.4}\\v=1.203\rm m/s[/tex]

The amplitude of wave is the half of the difference of highest point to the lowest point of the wave.

Now, the distance traveled by the boat  from its highest point to the lowest point is 0.700 m. Thus, the amplitude of each wave is,

[tex]A=\dfrac{0.7}{2}A=0.35\rm m[/tex]

Thus, the waves for the boat of fisherman is travelling with the speed of 1.203 m/s and the amplitude of each wave is 0.35 m.

Learn more about the speed of wave here;

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Answer:

C

Explanation:

Explanation:

Three balances can be written.

Mass of A into the column = mass of A out of the column

Mass of B into the column = mass of B out of the column

Mass of C into the column = mass of C out of the column

There are four unknowns: mass flow of the first feed stream, mass flow of the overhead stream, mass flow of the bottom stream, and percent B or C of the middle stream.

Since there are more unknowns than equations, there is not enough information to solve the equations.

Final answer:

The system described with three species allows for three material balances to be written, one for each chemical species A, B, and C.

Explanation:

The number of material balances that can be written on the overall system depends on the number of chemical species in the system. In the scenario provided, there are three species: A, B, and C. Therefore, three material balances can be written, one for each species.

The general form for a material balance is the sum of inputs minus the sum of outputs plus generation minus consumption equals accumulation. Since there is no indication of chemical reactions or accumulation (steady-state operation is assumed), the material balances for each species become:

For real-world applications, these material balances could be more complex, accounting for multiple units, reactions, and phase changes. However, with the information given and the assumption of steady-state, the total number of material balances that can be written for this system is three.