Define the following terms as used in thermodynamics giving examples of each; (i) Working fluid, (ii) Cycle, (iii) Intensive property and (iv) Reversible Process.

Answer:

H1 = 1/2 g t1^2      distance fallen in time t1

H2 = 1/2 g t2^2 = distance fallen in time t2

H2 / H1 = (t2 / t1)^2 = 4

One golf ball wlll fall 4 times the distance of the  other

Answer:

The lifeguard should run approximately 17.752 meters along the shore, before, jumping in the water

Explanation:

The given parameters are;

The rate at which the lifeguard runs = 5 m/s

The rate at which the lifeguard swims = 1 m/s

The horizontal distance of the child from the lifeguard = 30 meters along the shore

The vertical distance of the child from the lifeguard = 60 meters along the shore

Let x represent the distance the lifeguard runs

We have;

The distance the lifeguard swims = √((30 - x)² + 60²)

Time = Distance/Speed  

The time the lifeguard runs = x/5

The time the lifeguard swims = √((30 - x)² + 60²)/1

The total time = √((30 - x)² + 60²) + x/5

The minimum time is given by finding the derivative and equating the result to zero, as follows;

Using an online application, we have;

d(√((30 - x)² + 60²) + x/5)/dx = 1/5 - (30 - x)/(√((30 - x)² + 60²)) = 0

Which gives;

1/5 - (30 - x)/(√(x² - 60·x + 4500) = 0

(30 - x)/(√(x² - 60·x + 4500)) = 1/5

5×(30 - x) = √(x² - 60·x + 4500)

We square both sides to get;

(5×(30 - x))² = (x² - 60·x + 4500)

(5×(30 - x))² - (x² - 60·x + 4500) = 0

25·x² - 1500·x + 22500 - x² + 60·x - 4500 = 0

24·x² - 1440·x + 18000 = 0

Dividing n=by 24 gives;

24/24·x² - 1440/24·x + 18000/24 = 0

x² - 60·x + 750 = 0

By the quadratic formula, we have;

x = (60 ± √((-60)² - 4×1×750))/(2 × 1) =

Using an online application, we have;

x = (60 ± 10·√6)/(2)

x = 30 + 5·√6 or x = 30 - 5·√6

x ≈ 42.25 m and x ≈ 17.752 m

At x = 42.25

Time = √((30 - 42.247)² + 60²) + 42.247/5 ≈ 69.69 seconds

At x = 17.75

Time = √((30 - 17.752)² + 60²) + 17.752/5 ≈ 64.79 seconds

Therefore, the route with the shortest time is when the lifeguard runs approximately 17.752 meters (rounded to three decimal places) along the shore, before, diving in the water

The speed of the bullet was approximately 0.99 m/s.

Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance. Speed is the ratio of the distance travelled by time. The unit of speed in miles per hour.

We can start by using the conservation of momentum to relate the initial momentum of the bullet to the final momentum of the bullet-block system. If the bullet has mass m and initial speed v, then its initial momentum is p = mv.

After embedding in the block, the bullet-block system moves with the final speed vf. If we assume that the block is initially at rest, then the final momentum of the system is p' = (m + M)vf, where M is the mass of the block. By conservation of momentum, we have p = p', or

mv = (m + M)vf.

Solving for vf, we get

vf = mv / (m + M).

Now we can use the work-energy principle to relate the work done by friction to the kinetic energy of the bullet-block system. The work done by friction is given by

W = Fd = μmgd,

Where μ is the coefficient of kinetic friction, g is the acceleration due to gravity, and d is the distance the block slides. The kinetic energy of the bullet-block system is given by

K = (1/2)(m + M)vf².

By the work-energy principle, we have

W = K - 0,

Since the bullet-block system starts from rest. Substituting our expressions for W and vf, we get

μmgd = (1/2)(m + M)(mv / (m + M))²,

which simplifies to,

v = √(2μgd).

Plugging in the given values, we get

v = √(20.209.8 x 0.050) = 0.99 m/s.

Therefore, the speed of the bullet was approximately 0.99 m/s.

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Answer:

It obeys Newton's third law of motion.

Explanation:

The experiment obeys Newton's third law of motion because when the air in the balloon is released, the air moves in the downward direction which is an action of the balloon so in the reaction, the balloon moves in upward direction   so both action and reaction forces are equal in magnitude but moves in opposite direction which obeys Newton's third law of motion.

Answer:

147 newton.

Explanation:

Force is the product of mass and acceleration. The formula of force is given below: Force =  mass x acceleration, but due to height we put gravity instead of acceleration. So by putting values of mass i. e. 15 kg and gravity i. e. 9.8 meter/ second in the formula we get the net force of the ball which is 147 newton or 147 kg meter/second square.

Answer:

125 meters

Explanation:

we know that

v × t = d

v = 25

t = 5

so the distance should be 5 × 25

answer is 125 meters

Cause it's Corona Time

Given:

a.

b.

To find:

a. mg into km

b. cm into nm

Explanation:

a.

Hence,

b.

Hence,

Answer:

The answer is "26.208 km"

Explanation:

Given value:

\(\to S= 2.6 \ \frac{m}{s}\\\\\to t= 2.8 \ hours\)

Formula:

\(d= st\\\\d= 2.6 \times 2.8 \times \frac{60 \times 60}{1000}\\\\d= 26.208\ km\)

Explanation:

Boyle's law can be stated as the "volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant". It is mathematically expressed as;

    P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Charles's law states that the volume of a fixed mass of a gas varies directly as its absolute temperature if the pressure is constant.

It is mathematically expressed as;

            \(\frac{V1}{T1}\)    = \(\frac{V2}{T2}\)  

V and T are volume and temperature respectively

1 and 2 are the initial and final states

A. Different magmas generated in various places are isolated from one another. This assertion is untrue since magmas can interact with one another and mix.

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Answer:

An interaction between electricity and magnetism produces an electromagnet, while a permanent magnet induces the paper clip's magnetic field.

Explanation:

A p e x :)

Answer:

10 kg rock at 15 degrees

Explanation:

did test

The final temperature of the gas is 399.76 K when 5000 j of heat is added to two moles of an ideal monatomic gas, initially at a temperature of 500 k and the gas performs 7500 j of work.

The First Law of Thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W) by the system:

ΔU = Q - W

For an ideal monatomic gas, the change in internal energy (ΔU) can be calculated using the equation:

ΔU = (3/2) ×n× R ×ΔT, where n is the number of moles, R is the ideal gas constant (8.314 J/(mol K)), and ΔT is the change in temperature.

Given: Q = 5000 J (heat added)

W = 7500 J (work done)

n = 2 (number of moles)

Initial temperature (T1) = 500 K

Using the formula,

ΔU = Q - W,

(3/2) ×n× R ×ΔT = Q - W

(3/2) × 2 ×  8.314 × ΔT = 5000 - 7500

ΔT = -100.23 K

so the final temperature T2 = 500 - 100.23

T2 = 399.76 K

Therefore, the final temperature of the gas is 399.76 K.

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Answer:

I will answer in English, and i will give some representation for each case:

We have a = 6m and b = 8m.

If they are perpendicular, we have that:

a = (6m, 0) b = (0, 8m)

The adition is:

a + b = (6m, 8m).

If they are parallel, we have;

a = (6m, 0)

b = (8m, 0)

a + b = (6m + 8m, 0) = (14m, 0 )

If they are parallel but opposite:

a = (6m, 0)

b = (-8m, 0)

a + b = (6m - 8m,0) = (-2m, 0)

The magnitude of the force is approximately 0.0785 N. Since the currents in both wires are in the same direction, the force between the wires will be repulsive. Therefore, the correct answer is "The force on each wire will be repulsive."

The force between two parallel wires can be calculated using the formula:

F = (μ₀ * I₁ * I₂ * L) / (2πd)

Where F is the force, μ₀ is the permeability of free space (4π × 10^(-7) T m/A), I₁ and
I₂ are the currents in the wires,
L is the length of the wires, and d is the distance between the wires.

In this case, both wires carry the same current, I₁ = I₂ = 25 A, and the length of the wires is L = 25 m. The distance between the wires is d = 4.0 cm = 0.04 m.

Plugging these values into the formula, we can calculate the force:

F = (4π × 10^(-7) * 25 * 25 * 25) / (2π * 0.04) ≈ 0.0785 N

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Answer:

Character vs. self, Person vs. self, or conflict with the self (they're all the same thing.)

A drag car starts down a drag strip. If it can accelerate constantly at 17 m/s squared, then its speed after 5 seconds would be 85 meters/second.

There are three equations of motion given by  Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

v = u + at

  = 0 + 17×5

  = 85 m/s

S = ut + 1/2at²

200 = 0 + 0.5×5×t²

2.5t² = 200

t² = 200 / 2.5

t = 8.944 seconds

Thus, the speed of the car after 5 seconds would be 85 m/s and it would take 8.944 seconds to clear a 200-meter drag strip.

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Resolving power refers to the ability of a lens to distinguish between two closely spaced objects. It is determined by the wavelength of light and the numerical aperture of the lens.

Magnifying power, on the other hand, refers to the ability of a lens to enlarge the size of an object. It is determined by the focal length of the lens.
The term "parfocal" refers to a type of lens system where multiple lenses have the same focal point when the focus is adjusted. This means that when switching between different lenses, the focus remains the same, making it easier for the user to switch between lenses without losing focus.
Differentiating between the resolving power and magnifying power of a lens involves understanding their respective functions. Resolving power refers to the ability of a lens to distinguish between two closely spaced objects, or in other words, the clarity with which the lens can produce an image. Magnifying power, on the other hand, refers to the degree to which a lens can enlarge the image of an object.

The term "parfocal" is used to describe a set of lenses that, when interchanged on a microscope or other optical instrument, maintain their focus on the same object. This means that when you switch from one parfocal lens to another, only minimal adjustments to the focus are needed, allowing for a seamless transition between lenses with different magnifying powers.

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Resolving Power: It is the ability of a lens to separate or distinguish between closely spaced objects, reflecting the detail that can be seen with the lens.

Magnifying Power: It denotes how much larger an object appears through a lens compared to its actual size. High magnification doesn't necessarily mean better image quality.

Parfocal: This term refers to lenses that remain in focus even when the magnification or focal length changes. It enables swift adjustments in magnification without needing constant refocusing.

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1. With N = Newtons, m = meters, we have

\(F = -kx \iff \mathrm N = -k (\mathrm m) \implies k = \dfrac{\rm N}{\rm m}\)

N itself can be broken down according to Newton's second law,

\(F=ma \iff \mathrm N = \mathrm{kg} \, \dfrac{\rm m}{\rm s^2}\)

so the dimension of the force \(F\) is

\([F] = [M\cdot L\cdot T^{-2}]\)

(where M = mass, L = length, T = time) and hence the dimension of \(k\) is

\([k] = \left[\dfrac{M\cdot L\cdot T^{-2}}{L}\right] = \boxed{[M\cdot T^{-2}]}\)

2.

a. With N = Newtons, A = amperes, and m = meters,

\(F = \dfrac{k I_1 I_2}d \iff \mathrm N = \dfrac{k \,\mathrm A\,\mathrm A}{\rm m} \implies k = \boxed{\dfrac{\rm Nm}{\rm A^2}}\)

b. The dimension of \(k\) is

\([k] = \left[\dfrac{(M\cdot L\cdot T^{-2})\cdot L}{I^2}\right] = \boxed{[M\cdotL^2\cdot T^{-2}\cdot I^{-2}]}\)

(where I = current, with "I" as in capital i)

The bulb will light up in A but not in B because B is a short circuit.

When an electrical current bypasses the circuit's intended path or typical load and instead flows via an unanticipated path, the result is a short circuit. This could lead to a sudden surge in current that could harm electrical machinery, ignite a fire, or endanger nearby occupants' safety.

A direct connection between a power source's positive and negative terminals or a path with less resistance than the planned circuit are examples of the path that the current will take in a short circuit.

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The question addresses the stability of the atmosphere and the factors that determine convective stability or instability. It also explains the difference between the adiabatic lapse rate of dry air and wet saturated air.

a) The stability of the atmosphere is determined by the temperature profile and relative densities of the air cell and atmosphere. If the temperature of the surrounding atmosphere decreases with altitude at a rate greater than the adiabatic lapse rate of the air cell, the atmosphere is considered convectively stable.

In this case, the air cell will return to its original position. Conversely, if the temperature of the surrounding atmosphere decreases slower than the adiabatic lapse rate of the air cell, the atmosphere is convectively unstable. The air cell will continue moving in the same direction.

b) The adiabatic lapse rate refers to the rate at which temperature decreases with altitude for a parcel of air lifted or descending adiabatically (without exchanging heat with its surroundings). The adiabatic lapse rate of dry air is higher (around \(9.8^0C\) per kilometer) compared to the adiabatic lapse rate of wet saturated air (around 5°C per kilometer).

This difference arises because when water vapor condenses during the ascent of saturated air, latent heat is released, reducing the rate of temperature decrease. A diagram can illustrate the difference between the two lapse rates, showcasing their respective slopes.

c) When wet unsaturated air rises from the ocean surface, its temperature decreases at a rate equal to the dry adiabatic lapse rate. However, if the ambient lapse rate (temperature decrease with altitude) is higher than the adiabatic lapse rate for dry air, a temperature inversion layer forms at higher altitudes.

In this inversion layer, the temperature increases with altitude instead of decreasing. A schematic diagram can depict the temperature changes of the wet air in comparison to the ambient temperature, showing the inversion layer.

Cumulus clouds form at the altitude where the rising moist air reaches the level of the temperature inversion layer. These clouds are formed due to the condensation of water vapor as the air parcel cools to its dew point temperature.

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(a) To apply separation of variables, we assume T(x,t) = M(x)N(t), and substitute it into the heat conduction equation. This leads to two separate equations: d²M/dx² = -λ²M and dN/dt = -λ²/2N.

(b) For N(t), the ODE dN/dt = -λ²/2N has the general solution N(t) = \(Ce^{-\lambda^2t/2}\), where C is an arbitrary constant.

(c)  From the general expressions of M(x) and N(t), we combine them to define T(x,t) = (Acos(λx) + Bsin(λx). Hence, T(x,t) = \(5sin(4\pi x)e^(\lambda^2t/2)\).

(d) The arbitrary constant is determined as λ² = (4π)², which gives us λ = 4π. Thus, the final solution is T(x,t) = \(5sin(4\pi x)e^{-32\pi^2t}\).

a. To apply separation of variables, we assume T(x,t) = M(x)N(t), and substitute it into the heat conduction equation.

By differentiating twice with respect to x and once with respect to t, we obtain: ∂²M/∂x²N + 1/2M∂N/∂t = 0.

Since the left side depends on x and the right side depends on t, both sides must be equal to a constant, which we denote as -λ².

This leads to two separate equations:

d²M/dx² = -λ²M and dN/dt = -λ²/2N.

b. By solving the ODE for M(x), d²M/dx² = -λ²M,

we find that M(x) has the general form M(x) = Acos(λx) + Bsin(λx), where A and B are constants determined by the boundary conditions T(0,t) = T(3,t) = 0.

For N(t), the ODE dN/dt = -λ²/2N has the general solution:

N(t) = \(Ce^{-\lambda^2t/2}\), where C is an arbitrary constant.

c. From the general expressions of M(x) and N(t), we combine them to define:

T(x,t) = (Acos(λx) + Bsin(λx))\(Ce^{-\lambda^2t/2}\).

Simplifying the arbitrary constant C, we can rewrite it as

C = 5sin(4πx)/M(x) at t = 0,

which corresponds to the initial condition

T(x,0) = 5sin(4πx).

Hence, T(x,t) = \(5sin(4\pi x)e^{\lambda^2t/2}\).

d. Applying the boundary conditions T(0,t) = T(3,t) = 0,

we find that sin(λx) should be zero at x = 0 and x = 3.

This gives us λ = nπ/3, where n is an integer.

Substituting this value of λ into the expression for T(x,t), we obtain

T(x,t) = \(5sin(4\pi x)e^{-32\pi^2t}\),

which matches the given solution

T(x,t) = \(5sin(4\pi x)e^{-32\pi^2t}\).

Therefore, the arbitrary constant is determined as λ² = (4π)², which gives us λ = 4π.

Thus, the final solution is T(x,t) = \(5sin(4\pi x)e^{-32\pi^2t}\).

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Answer:

the first one i think

Explanation:

Answer:

The chemistries and technologies behind coin cells vary. Some are alkaline, others are lithium. Alkaline coin cell batteries have a nominal voltage of 1.5V. Lithium coin cell batteries, on the other hand, have a nominal voltage of 3V.

Answer:

deceleration? I think because the prefix de means to like not or to slow down

Answer:

here's the answers!

Explanation:

A mover carries a box up a flight of stairs.

A mover carries a box across a room.

A weightlifter lifts a barbell off the ground.

hope it helps u!

Answer:

A, D, E

Explanation:

Did it

Answer:

B is the answer!!

Explanation:

Answer:

Answer:

Dependent Variable is the answer