a package of mass m is released from rest at a warehouse loading dock and slides down the 2.0-m-high, frictionless chute to a waiting truck. unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (figure 1)
Part A
Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Part B
Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Express your answer to two significant figures and include the appropriate units.

Answer:

The velocity is  \(v_2= 0.45 \ m/s\)

Explanation:

From the question we are told that

      The initial speed of the hot water is  \(v_1 = 0.85 \ m/s\)

     The pressure from the heater  \(P_1 = 450 \ KPa = 450 *10^{3} \ Pa\)

      The height of the hot water before flowing is  \(h_1 = 0 \ m\)

      The height of bathtub above the heater is \(h_2 = 3.70 \ m\)

       The pressure in the pipe is \(P_2 = 414 KPa = 414 *10^{3} \ Pa\)

       The density of water is \(\rho = 1000 \ kg/m^3\)

Apply Bernoulli equation

      \(P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2\)

Substituting values

     \((450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )\)

=>   \(v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}\)

=>   \(v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}\)

=>    \(v_2= 0.45 \ m/s\)

As the fish tank is 20 inches by 12 inches by 12 inches, its volume is 47194744.32 mm³.

The space that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere.

1 inch = 0.0254 meters.

20 inches = 20 × 0.0254 meters = 0.508 meters = 508 mm.

12 inches = 12 × 0.0254 meters = 0.3048 meters = 304.8 mm.

Hence, The volume of the fish tank  = length × width × height

=  508 mm ×  304.8 mm ×  304.8 mm.

= 47194744.32 mm³

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The equation of motion for the rodent is x(t) = -1.12cos(3.20t), and the damping force is Fd = -0.62*vx(t). The damping force will cause the amplitude of the motion to decrease over time, and the rodent will eventually come to rest at the equilibrium position.

We can use the following equations to solve this problem:

F = -kx (Hooke's Law)

F = ma (Newton's Second Law)

a = d^2x/dt^2 (Definition of Acceleration)

Fd = -bv (Definition of Damping Force)

x(t) = A*cos(ωt + φ) (Equation of Motion for Simple Harmonic Motion)

We will need to use these equations to find the displacement, velocity, and acceleration of the rodent as a function of time, and then use that information to calculate the damping force and solve for the parameters of the motion.

First, let's find the natural frequency of the system:

ω = sqrt(k/m) = sqrt(3.50 N/m / 0.400 kg) = 3.20 rad/s

Next, let's assume that the rodent starts at its maximum displacement and moves in simple harmonic motion. We can use the equation of motion for simple harmonic motion to write:

x(t) = A*cos(ωt + φ)

where A is the amplitude of the motion and φ is the phase angle.

To find A and φ, we need to use the initial conditions. We know that at t=0, the rodent is at its maximum displacement, so x(0) = A. We also know that at t=0, the velocity of the rodent is zero, so vx(0) = -Aωsin(φ) = 0. This means that either A=0 (the rodent is not moving) or sin(φ) = 0 (the rodent is moving with maximum velocity). We will assume that the latter is true, so sin(φ) = 0 and cos(φ) = 1.

Now we can write:

x(t) = A*cos(ωt)

To find A, we use the fact that the rodent has a mass of 0.400 kg and is moving on a spring with force constant 3.50 N/m. The force on the rodent is given by:

F = -kx = -3.50 N/m * A*cos(ωt)

At maximum displacement, the force is equal to the weight of the rodent:

mg = 0.400 kg * 9.81 m/s^2 = 3.92 N

So we can write:

3.92 N = -3.50 N/m * A

A = -1.12 m

Therefore, the equation of motion for the rodent is:

x(t) = -1.12cos(3.20t)

To find the velocity and acceleration of the rodent, we take the derivative of the displacement with respect to time:

vx(t) = dx/dt = 3.58sin(3.20t)

ax(t) = d^2x/dt^2 = -11.46cos(3.20t)

To find the damping force, we use the equation:

Fd = -bv = -bdx/dt = -b3.58sin(3.20t)

We don't know the value of b, so we can't solve for it directly. However, we can use the fact that the damping force is equal to the work done by the damping force over one cycle of motion. This work is equal to the energy lost by the system due to damping. Since the system is losing energy at a rate proportional to its velocity, we can write:

Energy lost per cycle = Average damping force * Distance traveled per cycle

The distance traveled per cycle is equal to 2piA = 7.04 m, since the rodent moves from its maximum displacement to its minimum displacement and back again in one cycle.

The average damping force over one cycle is equal to the time average of the damping force:

<Fd> = (1/T)∫[0,T] -bdx/dt dt

where T = 2*pi/ω is the period of the motion. Evaluating the integral gives:

<Fd> = (1/T)∫[0,T] -b(-1.12)3.20sin(3.20*t) dt

<Fd> = 3.58*b

Since the energy lost per cycle is also equal to (1/2)kA^2, we can write:

(1/2)kA^2 = <Fd>2pi*A

Solving for b, we get:

b = (kA)/(2pi)

Substituting the given values, we get:

b = (3.50 N/m * 1.12 m)/(2*pi) = 0.62 Ns/m

Therefore, the equation of motion for the rodent is:

x(t) = -1.12cos(3.20t)

vx(t) = 3.58sin(3.20t)

ax(t) = -11.46cos(3.20t)

and the damping force is given by:

Fd = -0.62*vx(t)

Note that the negative sign indicates that the damping force acts in the opposite direction to the velocity of the rodent. This means that the damping force will cause the amplitude of the motion to decrease over time, and the rodent will eventually come to rest at the equilibrium position.

Therefore,The equation of motion for the rodent is x(t) = -1.12cos(3.20t), and the damping force is Fd = -0.62*vx(t).

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Answer:

The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the ball, u = 29.5 m/s

final velocity of the ball, v = -20.0 m/s (negative because it rebounds)

time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s

The force exerted on the brick wall by the ball is given as;

\(F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2\)

Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

Time period  and frequency of the pendulum are respectively 3.17 second and 0.315  second⁻¹. The velocity of its bob at the equilibrium is

140 cm/s.

Given parameters:

Length of the pendulum, L = 2.5 m = 250 cm.

Mass of the bob, m = 75 g

And,  bob is pulled 10 cm above equilibrium and then released.

acceleration due to gravity, g = 9.8 m/s².

Hence, time period of the pendulum is, T = 2π√(l/g)

= 2π√( 250/980) second.

= 3.17 second.

Frequency of the pendulum , n = 1/T = 1/3.17 second⁻¹ = 0.315  second⁻¹.

Let v be the velocity of the bob at equilibrium,

Kinetic energy at equilibrium = potential energy at the highest point

⇒ 1/2 × 75 × v² = 75 × 980 × 10

⇒ v = √(2×980×10)

= 140 cm/s.

So, the velocity at equilibrium after its bob is pulled 10 cm above equilibrium and then released is 140 cm/s.

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The distance travelled by the body is determined as 9 m.

The distance travelled by the body is calculated by applying the following kinematic equation as shown below.

Mathematically, the kinematic equation is given as;

v² = u² + 2as

where;

The given parameters include the following;

the initial velocity of the body = 0

the final velocity of the = 6 m/s

the constant acceleration of the body = 2 m/s²

the distance travelled by the body = ?

The distance travelled by the body is calculated as follows;

v² = u² + 2as

v² = 0 + 2as

v² =  2as

s = v² / 2a

s = ( 6² ) / ( 2 x 2 )

s = 9 m

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The complete question is below:

A body started from rest at constant acceleration of 2 m/s², the body reaches a velocity of 6 m/s. Calculate the distance travelled by the object to reach this velocity.

Answer: B

Explanation:

I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges

Answer:B :)

Explanation:

Answer:

Lever => \(d_{e} = d_{r}\)

Pulley => G = M x n (gravitational acceleration)

Wheel and axle => M.A = Radius of the wheel/radius of the axle = R/r

Inclined plane => It can be divided into two components: Fi = Fg * sinθ - parallel to inclined plane. Fn = Fg * cosθ - perpendicular one.

The five general principles from the APA are meant to: C. Guide and inspire good conduct.

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

There are 90 phones of volume, 10-7 W/m2 of sound intensity, and 0.0314 watts of source power.

A straightforward frequency analysis compares the values of the fields you provide and generates a report listing each value for those fields along with the frequency at which each value occurs.

The rate at which a sound power wave repeats itself, also known as frequency or pitch, is measured in cycles per second. Bullfrog calls and cricket chirps have lower frequencies than drum beats and whistles, respectively.

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Answer:

True

Nitrogen family elements consists of atoms having 5 electrons in their outer energy level.

Please mark me the brainliest

Hope it may helps you

The mass of the wire of lowest A on a piano is 0.00165 kg.

The frequency of a vibrating string is given by the equation:

f = (1/2L) * sqrt(T/μ)

where f is the frequency of the string, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string (mass per unit length).

We know the frequency of the lowest A on a piano is 27.5 Hz. We also know that one half wavelength occupies the string, so the length of the string is half the wavelength:

L = (1/2) * λ

The wavelength of a sound wave is given by:

λ = 2L/n

where n is the number of nodes (points of zero displacement) in the wave. For the lowest A on a piano, n = 1, so we can write:

λ = 2L

Substituting this into the equation above for L, we obtain:

L = λ/2

Now we can substitute these values into the first equation:

27.5 = (1/2)(λ/2) * sqrt(308/μ)

Simplifying, we get:

λ = 4L

308/μ = 4(27.5)^2 (1/4)

μ = 0.000824 kg/m.

Since μ = m/L, where m is the mass of the wire and L is its length, we can find the mass of the wire by multiplying the linear mass density by the length of the string:

m = μL

The length of the string is given as 2.00 m, so we can write:

m = 0.000824 kg/m * 2.00 m = 0.00165 kg

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Answer:

B. space quantization.

Explanation:

In 1921, Otto Stern developed the idea behind this experiment, while Walther Gerlach performed the actual experiment in 1922. The Ster-Gerlach experiment provides prove to the fact that the spatial orientation of angular momentum is quantized. To demonstrate the experiment, silver atoms were made to travel through a magnetic field path.

Before they hit the screen(usually a glass slide), they were deflected because of their non-zero magnetic moment. There was an expected result for this experiment, but the actual observation on the glass slide was a continuous distribution of the silver atoms that actually hit the glass. This experiment was useful in proving that in all atomic-scale systems, there was a quantization of angular momentum.

Answer

To get critical pressure

We use

Pc = a/(27b²)

So

= 3.610/(27 X 0.0429²)

We have

= 72.7 atm

Critical temperaturewe

We use

Tc = 8a/27Rb

= 8 x 3.610/(27 x 0.0812 x 0.0429)

= 307 K

Critical volume

We use

Vc =3b =

3 x 0.0429

= 0.129L/mol

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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The magnitude of the acceleration due to gravity of the planet is 11.06 m/s², if the pendulum completes 101 full swing cycles in 136 seconds.

Length of the pendulum, l = 51.0 cm = 0.51 m

Time taken to complete 101 full swings cycle is 136 seconds, so

Time period of the pendulum, T = 136/101 = 1.35 sec

Let the value of gravitational acceleration on the planet, = gₓ

We know the equation of simple pendulum is: T = 2π√(l/g)

1.35 = 2π√(0.51/g)

Squaring of both sides,

1.35² = (2π)²×(0.51/gₓ)

1.82 = 20.13/gₓ

gₓ = 20.13/1.82

gₓ = 11.06 m/s²

--The given question is incorrect, the correct question is:

"after landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 51.0 cm . the explorer finds that the pendulum completes 101 full swing cycles in a time of 136 s. What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second."--

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Heat emitted from a candle, Microwaves from a microwave oven, Radio waves transmitted between towers, Sound waves from a car radio.

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It has the greatest Potential energy when the ball reaches its highest point.The correct option is D

Potential energy is the energy that an object possesses due to its position or state, which can be converted into work or kinetic energy when acted upon by a force.

There are three types of Potential Energy such as :

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Answer:

1) PE=mgh

2)when the ball reaches its highest point

3)21,599,200 J

4) The object’s mass, gravity, and height determine its potential energy.

5)Members of a team work together to create a plan, and then members choose different tasks to carry out the plan.

Answer:

hey iam bored pls join

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Password: 1111

The term change refers to an alteration of present conditions.

The term change refers to an alteration of present conditions. When a change occurs, things cease to be the way they were. As a student, the complexity of academic work increases with time. This is a change that has taken place.

As an employee, you are given real world tasks to accomplish rather than mere mental exercises like when you were a student, this is change that has taken place.

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The nucleons have less mass, because matter is converted into binding energy. Option D is correct.

During the process of combining a neutron and a proton to form a nucleus, a small amount of mass is converted into binding energy. This is due to the strong nuclear force that holds the nucleus together. The mass of the nucleus is slightly less than the sum of the masses of the individual nucleons, and the difference in mass is referred to as the mass defect.

This mass defect is related to the binding energy of the nucleus through Einstein's famous equation E=mc², where E is energy, m is mass, and c is the speed of light. The mass defect represents the amount of mass that is converted into binding energy to hold the nucleus together. Option D is correct.

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Given:

• Length of wire = 3.0 cm

• Current = 10 A

• Magnetic field = 0.27 T

Let's find the magnetic force on the wire.

To find the magnetic force on the wire, apply the formula:

Where:

B is the magnetic field = 0.27 T

I is the current = 10 A

L is the length of wire in meters = 3.0 cm = 0.03 m

Since it is perpendicular, θ = 90 degrees.

F is the magnetic force.

Substitute the values into the formula and solve for F.

We have:

Therefore, the magnetic force on the wire is 8.1 x 10⁻² N.

ANSWER:

8.1 x 10⁻² N

a. Aster is 56.3 m at 3.16° north-east from her initial position

b. She has to head to 183.16° or 86.84° south of west to return to her initial position

Let Aster's initial position be represented by the vector r = 0i + 0j

Since she then walks walks first 70 m in the direction 37° north of east, let this displacement be represented by the vector u = (70sin37°)i + (70cos37°)j

=  (70 0.6018)i + (70 0.7986)j

= 42.13i + 55.9j

Also, she then walks  82 m in the direction 20° south of east. Let this displacement be represented by the vector v = (82sin20°)i - (82cos20°)j

= -(82 0.3420)i + (82 0.9397)j

= 28.05i - 77.05j

Finally, she walks 28 m in the direction 30° west of north. Let this displacement be represented by the vector, w = -(28sin30°)i + (28cos30°)j = -(28 0.5)i + (28 0.8660)j

= -14i + 24.25j m

So, the total displacement is R = r + u + v + w

= 42.13i + 55.9j + 28.05i + (-77.05)j +  (-14)i + 24.25j m

= 56.18i + 3.1j

So, how far she walks is the magnitude of R. The magnitude of a vector Z = xi + yj is Z = √(x² + y²)

So, the magnitude of R = √((56.18)² + (3.1)²)

= √(3156.19 + 9.61)

= √3165.8

= 56.3 m

The direction of a vector Z = xi + yj is given by Ф = tan⁻¹ (y/x)

So, the direction of R is Ф' = tan⁻¹ (3.1/56.18)

= tan⁻¹ (0.0552)

= 3.16°

So, Aster is 56.3 m at 3.16° north-east from her initial position

To return to her original position, the displacement vector is V = r - R

= 0i + 0j - (56.18i + 3.1j)

= -56.18i - 3.1j

So, the direction of V is Ф" = tan⁻¹ (-3.1/-56.18)

= tan⁻¹ (0.0552)

= 3.16°

Since this is in the third quadrant, we have that the direction she must go to return to her original position is α = 180° + 3.16°

= 183.16°

or 90° - 3.16°

= 86.84° south of west

So, she has to head to 183.16° or 86.84° south of west to return to her initial position

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Answer:

4.408 \(\mathsf{M_{sun}}\)

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

\((m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}\)

\(\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}\)

\(\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}\)

combined mass (m₁+m₂)  = 4.408 \(\mathsf{M_{sun}}\)

Answer:

Explanation:

From the given information if we assume that the runner didn't  come back to her original position.

Her initial velocity = 0 m/s

Final Velocity = 9.6 m/s

Time Taken = 2 seconds

She started from rest therefore her initial time would be 0 seconds.

We know that,

Average acceleration = (final velocity - initial velocity)/(final time - initial time)

=(v-u)/t1-t0

=(9.6 - 0)/(2-0) m/s^2

=4.8 m/s^2

To calculate the resultant force on the aircraft as it accelerates from the catapult, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

Since all of the kinetic energy at take-off is from the work done on the aircraft by the catapult, we can equate the work done by the catapult to the kinetic energy of the aircraft at take-off, that is:

Work done by catapult = kinetic energy of aircraft at take-off

The work done by the catapult is given by the force provided by the catapult multiplied by the distance over which it acts, that is:

Work done by catapult = force × distance

The distance over which the force provided by the catapult acts is given as 150 m, so we have:

Work done by catapult = force × 150

The kinetic energy of the aircraft at take-off is given by:

(1/2) × mass × velocity^2

Since the aircraft is initially at rest, its initial velocity is zero, so we have:

kinetic energy of aircraft at take-off = 0.5 × mass × (final velocity)^2

Using the work-energy principle, we can equate the two expressions for work done and kinetic energy, that is:

force × 150 = 0.5 × mass × (final velocity)^2

Solving for force, we get:

force = 0.5 × mass × (final velocity)^2 / 150

Therefore, the resultant force on the aircraft as it accelerates is given by:

force = 0.5 × mass × (final velocity)^2 / 150

Note that we need to know the mass and final velocity of the aircraft in order to calculate the resultant force.

Taking half of the given length and pulling it to a length of 2L as before, the frequency obtained by vibrating in the basic method is 6fo. option(ii)

When a bow vibrates in its fundamental mode with a length of 2L, the frequency is denoted as f. Now, let's consider the scenario where half of the given length (L) is pulled to a length of 2L.

In the fundamental mode of vibration, the frequency is inversely proportional to the length of the vibrating object. Therefore, if the length of the bow is halved to L, the frequency would double to 2f.

When this new length of L is pulled to a length of 2L, we need to determine the frequency obtained in the fundamental mode.

Since the frequency is inversely proportional to the length, we can use the inverse relationship to find the new frequency.

If the original frequency was 2f at length 2L, when the length is reduced to L, the new frequency would be (2f)/(2L) = f/L.

Now, if this length of L is stretched to 2L again, the new frequency in the fundamental mode would be (f/L) * (2L) = 2f. option(ii)

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