if no corrective action is taken by the pilot as angle of bank is increased, how is the vertical component of lift and sink rate affected?

The average rate of production of the aerogel is 50,000 grams per minute.

Aerogels have been investigated for a variety of uses, including catalysis, thermal insulators, solar energy applications, piezoelectric, energy conversions-storage, low-temperature glass creation, sensors, adsorption, and photocatalysis. They are porous and have low density structures.

To convert from cubic meters (m³) to cubic centimeters (cm³), we need to multiply by 1,000,000,000.

0.360 m³ = 0.360 × 1,000,000,000 = 360,000,000 cm³

So the company can produce 360,000,000 cm³ of aerogel in 24 hours.

Let's assume it's 0.2 g/cm³ (this is a typical value for silica aerogel).

Mass = density x volume

Mass = 0.2 g/cm³ x 360,000,000 cm³

Mass = 72,000,000 grams

The company can produce 72,000,000 grams of aerogel in 24 hours.

To find the average rate of production in grams per minute, we can divide the total amount produced by the number of minutes in 24 hours (24 x 60 = 1440 minutes).

Average rate of production = 72,000,000 grams / 1440 minutes

Average rate of production = 50,000 grams/minute (to the nearest thousand)

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Answer:

\(30 \div 33750 = 0.008888\)

Given that, 1 Liter equals 1.0567 quartz and 1 quartz equals 32 oz, the volume 1.00 oz equals 0.02957 L.

Given that;

First, we convert 1.00 oz to qt

32 oz = 1 qt

1.00 oz = x

x = 1.00 / 32

x = 0.03125 qt

Now, we convert 0.03125 qt to liter

1.0567 qt = 1 L

0.03125 qt = x

x = 0.03125 / 1.0567

x = 0.02957 L

Given that, 1 Liter equals 1.0567 quartz and 1 quartz equals 32 oz, the volume 1.00 oz equals 0.02957 L.

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a. The angular speed of the rotating nylon line is 314.16 radians per minute.

b. The linear speed of a point on the tip of the line is approximately 2617 inches per minute.

The angular speed of the rotating nylon line can be determined by converting the rotational speed from revolutions per minute (rpm) to radians per minute.

Since one revolution is equal to 2π radians, the angular speed is calculated as 3000 rpm multiplied by 2π, resulting in approximately 314.16 radians per minute.

To find the linear speed of a point on the tip of the line, we need to multiply the angular speed by the radius of the circular path. Given that the line is 5 inches long, the radius is half of that, or 2.5 inches.

The linear speed can be obtained by multiplying the angular speed of 314.16 radians per minute by the radius of 2.5 inches, resulting in approximately 785.4 inches per minute.

Rounding to the nearest inch per minute, the linear speed of a point on the tip of the line is approximately 2617 inches per minute.

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Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

Answer:

-3m+7m =  4m

Explanation:

As he walks south, he is going down 3m (-3m). Then he walks up 7m (+7m).

You subtract the final position from the initial position to get displacement.

7m - 3m = 4m

Explanation:

divide the distance by the time it takes to travel that same distance, then add your direction to it.

Or change in displacement divided by change in time.

The value of theta for the maximum height is 30°.

Calculation:-

Since the angle from the horizontal is Cosθ, therefore the angle from the vertical is sinθ.

Hmax = U² sin²θ/2g

11.25 = 30² sin²θ/2g

sin²θ = (11.25 × 2 × 10)/900

   ⇒    sin²θ   =   0.25

    ⇒   sin θ = 0.5

    ⇒   sin θ = 1/2 = 30°

The horizontal pace of a projectile is regular (in no way converting in value), and there's a vertical acceleration because of gravity; its cost is 9.8 m/s/s, down, The vertical speed of a projectile changes via nine.8 m/s every second, The horizontal motion of a projectile is impartial of its vertical movement.

Projectile motion is a form of motion experienced with the aid of an object or particle that is projected in a gravitational field, such as from Earth's floor, and movements alongside a curved route below the action of gravity best.

Projectile motion is the movement of an object thrown (projected) into the air. After the initial force that launches the item, it most effectively reports the pressure of gravity. The object is known as a projectile, and its route is called its trajectory.

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Answer:

All forces are either balanced or unbalanced and each force acts in different ways.

The  pressure on the upper wing surface is 4.10 x 10^3 Pa.

The lift force on an airplane wing is due to the pressure difference between the upper and lower surfaces of the wing. According to Bernoulli's principle, the pressure of a fluid decreases as its speed increases. Therefore, the air moving over the curved upper surface of the wing must have a lower pressure than the air moving over the flat lower surface of the wing.

The lift force can be calculated using the formula:

L = Cl * ρ * A * v^2 / 2

where L is the lift force, Cl is the lift coefficient, ρ is the density of the air, A is the area of the wing, and v is the speed of the air.

Since the airplane is in level flight, the lift force is equal to the weight of the airplane:

L = mg

where m is the mass of the airplane and g is the acceleration due to gravity.

We can use these two equations to find the speed of the air over the wing:

mg = Cl * ρ * A * v^2 / 2

v^2 = 2mg / (Cl * ρ * A)

v = sqrt(2mg / (Cl * ρ * A))

Let's assume that the lift coefficient is the same for both the upper and lower surfaces of the wing. Then, the lift force on each wing is:

L/2 = Cl * ρ * A * v^2 / 4

The pressure difference between the upper and lower surfaces of the wing can be found using the formula:

ΔP = 2 * (L/2) / A

Substituting the expressions for L/2 and v^2, we get:

ΔP = Cl * ρ * v^2

ΔP = Cl * ρ * (2mg / (Cl * ρ * A))

ΔP = 2mg / A

Substituting the given values, we get:

ΔP = 2 * (1.10 x 10^4 kg) * 9.81 m/s^2 / (39.0 m^2)

ΔP = 5.59 x 10^4 Pa

Since the pressure on the lower wing surface is given as 6.00 x 10^4 Pa, the pressure on the upper wing surface is:

Pupper = Plower - ΔP

Pupper = (6.00 x 10^4 Pa) - (5.59 x 10^4 Pa)

Pupper = 4.10 x 10^3 Pa

Therefore, the pressure on the upper wing surface is 4.10 x 10^3 Pa.

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It is correctly interpreted as the Gravitational redshift .

Albert Einstein evolved the overall concept of relativity from 1907 to 1915. the overall idea of relativity is also known as the principle of gravitation. in line with this principle, the gravitational pressure among two of our bodies absolutely depends on their hundreds and on the distance between two bodies.

He additionally observed that time and area are interlinked to each other and are known as space-time. Through the equation of the overall principle of relativity, he found that huge objects brought on a distortion in area-time. In 1915, Einstein derived the field equation associated with the curvature of spacetime with the electricity, mass, and any momentum inside it.

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The maximum velocity during this interval of time is approximately 20 m/s, and it occurs at t ≈ 3 + √10 seconds.

To find the maximum velocity and the time at which it occurs, we need to differentiate the position function with respect to time.

Given: s = -2/3 t³ + 6t² + 2t

(a) To find the maximum velocity, we differentiate the position function to get the velocity function: v = ds/dt

Taking the derivative of the position function, we have:

v = d/dt (-2/3 t³ + 6t² + 2t)

v = -2t² + 12t + 2

To find the maximum velocity, we set the derivative equal to zero and solve for t: -2t² + 12t + 2 = 0

We can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

Substituting the values into the formula, we get:
t = (-12 ± √(12² - 4(-2)(2))) / (2(-2))

t = (-12 ± √(144 + 16)) / (-4)

t = (-12 ± √160) / (-4)

t = (-12 ± 4√10) / (-4)

t = 3 ± √10

Since we're looking for the time during the interval from t = 0 to t = 6 s, we discard the negative root: t = 3 + √10

So, the maximum velocity occurs at t = 3 + √10 seconds.

To find the maximum velocity, substitute the value of t into the velocity function: v = -2(3 + √10)² + 12(3 + √10) + 2

v ≈ 20 m/s

Therefore, the maximum velocity during this interval of time is approximately 20 m/s, and it occurs at t ≈ 3 + √10 seconds.

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The average angular velocity of the lathe is 3.75 rad/s.

First, we can find the final angular velocity of the lathe at the end of the constant velocity phase by using the kinematic equation:

ωf = ωi + αt

where ωi is the initial angular velocity (which is zero in this case), α is the angular acceleration (\(0.60 rad/s^2)\), and t is the time it takes to reach the constant velocity (10 s):

ωf = 0 + 0.60 rad/s^2 * 10 s = 6.0 rad/s

Next, we can use the constant velocity phase to find the total angle turned by the lathe during this period, which is given by:

θ2 = ω * t2

where ω is the constant angular velocity during this period (which is also 6.0 rad/s) and t2 is the time period (20 s):

θ2 = 6.0 rad/s * 20 s = 120 rad

Finally, we can use the deceleration phase to find the total angle turned by the lathe during this period, which is given by:

θ3 = ωf * t3 + (1/2) * (-α) * t3^2

where ωf is the final angular velocity (6.0 rad/s), α is the angular deceleration (-0.60 rad/s^2), and t3 is the time period (10 s):

θ3 = 6.0 rad/s * 10 s + (1/2) * (-0.60 rad/s^2) * (10 s)^2 = 30 rad

The total angle turned by the lathe is therefore:

Δθ = θ1 + θ2 + θ3 = 0 + 120 rad + 30 rad = 150 rad

The total time taken by the lathe is:

Δt = t1 + t2 + t3 = 10 s + 20 s + 10 s = 40 s

Therefore, the average angular velocity of the lathe is:

ω_avg = Δθ / Δt = 150 rad / 40 s = 3.75 rad/s

So the average angular velocity of the lathe is 3.75 rad/s.

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Full  Question: A lathe, initially at rest, accelerates at 0.60 rad/s^2 for 10 s, then runs at a constant angular velocity for 20 s, and finally decelerates uniformly for 10 s to come to a complete stop. What is its average angular velocity?

The intensity at a point on the screen 2.00 mm from the center of the central maximum is approximately 0.034 W/m². The intensity at a point on the screen 1.50 mm from the center of the central maximum is approximately 0.024 W/m².

I = Imax cos² (πd sin θ / λ),

where Imax is the intensity at the center of the interference pattern, d is the distance between the two slits, θ is the angle between the line connecting the point on the screen to the center of the interference pattern and the line perpendicular to the screen, and λ is the wavelength of the light.

To find the angle θ, we can use the small angle approximation:

sin θ ≈ θ ≈ y/L,

where y is the distance from the center of the interference pattern to the point on the screen, and L is the distance between the slits and the screen.

We are given d = 0.0720 mm, λ = unknown, L = 0.800 m, Imax = 0.0700 W/m², and the distance from the center of the central maximum to the first minimum y = 3.00 mm.

Using the given distance y, we can find the value of sin θ:

y/L = sin θ,

3.00 mm / 0.800 m = sin θ,

sin θ = 0.00375.

Now we can solve for the wavelength λ:

Imax cos² (πd sin θ / λ) = I,

0.0700 W/m² cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I,

cos² (π(0.0720 × 10⁻³ m)(0.00375) / λ) = I / 0.0700 W/m²,

π(0.0720 × 10⁻³ m)(0.00375) / λ = ± cos⁻¹ (√(I / 0.0700 W/m²)),

λ = π(0.0720 × 10⁻³ m)(0.00375) / cos⁻¹√(I / 0.0700 W/m²)),

λ = 5.70 × 10⁻⁷ m (for the positive root).

Now we can find the intensities at the given distances from the center of the central maximum.

For y = 2.00 mm:

sin θ = y/L = 2.00 mm / 0.800 m = 0.00250,

I = Imax cos² (πd sin θ / λ)

I = 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.00250) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².

So the intensity at a point on the screen 2.00 mm from the center of the central maximum would be approximately 0.034 W/m².

For y = 1.50 mm:

sin θ = y/L = 1.50 mm / 0.800 m = 0.001875,

I = Imax cos² (πd sin θ / λ)

I= 0.0700 W/m² cos² (π(0.0720 × 10⁻³m)(0.001875) / (5.70 × 10⁻⁷ m))² ≈ 0.034 W/m².

I ≈ 0.024 W/m².

So the intensity at a point on the screen 1.50 mm from the center of the central maximum would be approximately 0.024 W/m².

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Equation (3) is a true equation for an LR circuit because it represents Kirchhoff's voltage law, which states that the sum of the voltage drops across all elements in a closed loop is equal to the applied voltage.

Equation (8) is a true equation for an LC circuit because it represents Kirchhoff's current law, which states that the sum of the currents entering a node is equal to the sum of the currents leaving the node.

In an LR circuit, equation (3) is true because it represents the balance between the applied voltage (V) and the voltage drops across the inductor (L) and resistor (R) in a closed loop. The voltage drop across the inductor (L) is given by L(di/dt), where di/dt represents the rate of change of current. The voltage drop across the resistor (R) is simply IR, where I is the current flowing through the circuit. Summing these voltage drops gives V = L(di/dt) + IR, which is equation (3).

In an LC circuit, equation (8) is true because it represents the balance of currents at a node. At any given instant, the current flowing into the node splits into two branches: one through the capacitor (C) and the other through the inductor (L). The current through the capacitor is given by C(dv/dt), where dv/dt represents the rate of change of voltage. The current through the inductor is simply IL, where L is the inductance. Summing these currents at the node gives IL + C(dv/dt) = 0, which is equation (8).

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Answer:

4 m/s in negative acceleration

Explanation:

Acceleration = V- U/t

Where V is the final velocity

U is the initial velocity and t is the time given.

U = 65 m/s

V= 25 m/s

T= 10 seconds

Acceleration= (25m/s - 65m/s)÷10secs

= - 40/10

= -4m/s^2

Hence, it has a negative acceleration.

Answer:

A. negative acceleration of 4 m/s2

Hope this helps!

Explanation:

When parked on the highway at night, cars should use their hazard lights or emergency flashers.

When parked on the highway at night, cars should use their hazard lights or emergency flashers. These lights are designed to alert other drivers of a potential hazard or obstruction on the road. By activating the hazard lights, the parked car becomes more visible to oncoming traffic, reducing the risk of accidents.

Hazard lights typically consist of a pair of high-intensity, blinking lights located at the front and rear of the vehicle. They emit a bright, attention-grabbing signal that can be easily seen from a distance, even in dark or adverse weather conditions.

The blinking pattern of the lights distinguishes them from the regular headlights or taillights of moving vehicles, indicating that the car is stationary and that caution should be exercised.

Using hazard lights while parked on the highway helps to warn approaching drivers to slow down and proceed with caution. It also helps to prevent rear-end collisions or other accidents caused by drivers failing to notice the stationary vehicle in time.

However, it is important to note that hazard lights should only be used when the car is parked in a safe location off the road and not obstructing traffic flow.

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a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

K.E = 100%P.E - 18%P.E

where;

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

v² = 0.804

v = √0.804

v = 0.9 m/s

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

P.E = mgh

h = P.E/mg

h = (0.374)/(1 x 9.8)

h = 0.038 m

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

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Answer:

1,860

Explanation:

The acceleration of the car is \(11.63 m/s^2.\) The force produced by the car is 4,876.4 Newtons.

To find the acceleration of the car, we can use the formula:

acceleration = (final velocity - initial velocity) / time

In this case, the initial velocity is 0 m/s, the final velocity is 50 m/s, and the time is 4.3 s. Substituting these values into the formula, we get:

acceleration = (50 m/s - 0 m/s) / 4.3 s =\(11.63 m/s^2\)

Therefore, the acceleration of the car is\(11.63 m/s^2.\)

To find the force produced by the car, we can use Newton's second law of motion, which states that force is equal to mass times acceleration:

force = mass x acceleration

In this case, the mass of the car is 420 kg, and the acceleration is 11.63 m/s^2. Substituting these values into the formula, we get:

\(force = 420 kg \times 11.63 m/s^2 = 4,876.4 N\)

Therefore, the force produced by the car is 4,876.4 Newtons.

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Answer:

variable that doesnt change

Explanation:

contolled variable is that variable that does not change throughout the course of the experiment

Answer:

The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object.

Answer:

1700 Joules

Explanation:

Work=force x distance

Force = 170 kg

Distance= 10 Meters

170 x 10 = 1700 Joules of work

The sound level of 86dB when fired simultaneously will have sound level of 82.9898dB.

Theory-

According to the Beta factor model, the common cause will have an equal impact on every member of a component group that shares a common cause.

The beta factor model is simple to use, models common cause failures, and only requires one parameter to be determined.

A drawback of the beta factor model is that it is impossible to quantify the failure of k–m components within a common cause component group.

As a result of multiplying the partial beta factors, the beta factor is calculated.

Mathematics-

\(\beta =10log\frac{2I}{i} \\\\\beta 1=10log\frac{I}{i} \\\\\beta =10log2 +\beta 1\\\\\beta 1=\beta -10log2\)

Calculation-

\(\beta 1=86-10log2\\\\\beta 1=86-3.0102\\\\\beta 1= 82.9898dB\)

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What is their to fix?

Answer:

4x10=30

Explanation:

You can easily know it's multiplication when you see the sign. It gives you the number but you have to plug it in. So once you plug in the numbers then you should get either 4x10 or 10x4 which means both is equal to 40