An ultracentrifuge spins biological samples at very high speeds in order to separate their constituent microscopic components. One such instrument spins at 120,000 rpm, and the samples experience a centripetal acceleration of 6.6×10^6 m/s^2. What is the radius of the circle around which the samples travel?

Answer with Explanations:

Given:

mass of block, m = 0.244 kg

spring constant, k = 5075 N/m

compression distance in spring, x = 0.106 m

Find how high mass reaches after release (measured from point of release)

Solution:

Use conservation of energy, assuming a light spring (of negligible mass)

stored spring potential energy = gravitational potential energy of block

PE1 = PE2

PE1 = kx^2/2 = 5075*0.106^2/2 = 28.512 J

PE2 = mgh = 0.244 * 9.8 * h = 2.3912 h J

Solve for h = 28.512 / 2.3912 = 11.923 m

To find the maximum height a block rises after being released from a compressed spring, we use the conservation of energy principle to equate the initial spring potential energy to the gravitational potential energy at the peak height, then solve for that height.

The problem involves applying the conservation of energy principle to a mass-spring system when the block is released from compression. Initially, all the energy is stored as potential energy in the compressed spring (spring potential energy), and when the spring is uncompressed, that energy will be converted into kinetic energy of the mass. Finally, when the block leaves the spring and moves upward, its kinetic energy will be converted into gravitational potential energy until the block comes to a stop at its maximum height.

Let's examine the steps needed to solve the problem:

Let's perform these calculations for the provided problem:

Answer:

r² / 110² − 3z² / 1375² = 1

Explanation:

The equation of a hyperboloid (which is a hyperbola rotated about the z axis or conjugate axis) that is centered at the origin is:

x² / a² + y² / b² − z² / c² = 1

If the cross sections are circular rather than elliptical, then a = b.

(x² + y²) / a² − z² / c² = 1

Or, if you prefer cylindrical coordinates:

r² / a² − z² / c² = 1

We know that at z = 0, r = 110.  And at z = -500, r = 130.

110² / a² − 0 = 1

130² / a² − (-500)² / c² = 1

Solving:

a² = 110²

c² = 1375² / 3

Plugging in:

r² / 110² − 3z² / 1375² = 1

To find an integer solution to given linear equations, iterate through integer values for x and y within the range of -10 to 10 and substitute them into the equations to check for solutions.

To find an integer solution for the set of given linear equations using brute force, you would substitute integer values for x and y within the specified range of -10 to 10. The equations given as examples, such as 7y = 6x + 8, 4y = 8, and y + 7 = 3x, are all linear equations that can be rearranged into the standard form y = mx + b, where m is the slope and b is the y-intercept.

In practice, you would start with -10 for x, solve for y in each equation, and check if y is an integer. Repeat incrementally until you reach 10 for x. If you find a pair of x and y values that satisfy both equations, you've found a solution. Remember to verify your solution by substiting the found values back into the original equations to ensure they hold true.

For example, if the equations were 2x + 3y = 6 and 4x - y = 10, you would try each pair of x and y between -10 and 10 until you find a pair that works for both equations. This brute force method is not the most efficient, but it will eventually yield an integer solution, if one exists within the given range.

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The solution that satisfies both equations is x = 3 and y = 2.

In many engineering and scientific applications, finding solutions to systems of linear equations is crucial. Given the linear equations 8x + 7y = 38 and 3x - 5y = -1, we are tasked with finding integer solutions for x and y in the range of -10 to 10.

Steps to Solve Using Brute Force :

We will systematically check all integer pairs (x, y) within the specified range:

After performing these steps, we find the solution: x = 3 and y = 2. Both equations are satisfied by this pair because:

8(3) + 7(2) = 24 + 14 = 38 and 3(3) - 5(2) = 9 - 10 = -1

Thus, x = 3 and y = 2 is the correct solution.

To find the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia is 5.10 ✕ 10^8 kg · m^2 and the crew experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia increases, resulting in a smaller apparent acceleration for those remaining at the rim.

To determine the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia of the system is 5.10 ✕ 10^8 kg · m^2 and the crew on the rim experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia of the system increases. Therefore, the angular speed decreases, resulting in a smaller apparent acceleration for those remaining at the rim.

We can use the conservation of angular momentum equation:

I₁ω₁ = I₂ω₂

Where:

Substituting the given values:

(5.10 ✕ 10^8 kg · m^2)(ω₁) = (5.10 ✕ 10^8 kg · m^2 + (100)(65.0 kg)(151 m)²)(ω₂)

Simplifying and solving for ω₂:

ω₂ = (5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]

To find the apparent acceleration experienced by the managers remaining at the rim, we can use the centripetal acceleration formula:

a = rω²

Where:

Substituting the given values:

a = (151 m)(ω₂)²

Substituting the expression for ω₂ obtained previously:

a = (151 m)[(5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]]²

Simplifying and solving for a:

a = (151 m)(5.10 ✕ 10^8 kg · m^2)²(ω₁)² / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]²

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The new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].

Let's break down the problem step by step to find the apparent acceleration experienced by the managers remaining at the rim after 100 people move to the center of the space station.

1. **Initial Moment of Inertia:**

  [tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]

  [tex]\[I_{\text{rim}} = (150 \times 65.0 \, \text{kg}) \times (151 \, \text{m})^2\][/tex]

 [tex]\[I_{\text{100 moved}} = (100 \times 65.0 \, \text{kg}) \times (0 \, \text{m})^2\][/tex]

  [tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]

2. **Conservation of Angular Momentum:**

  [tex]\(I_1 \omega_1 = I_2 \omega_2\)[/tex]

3. **Calculate Angular Speeds:**

  Given [tex]\(I_1 = 5.10 \times 10^8 \, \text{kg} \cdot \text{m}^2\),[/tex]

  solve for [tex]\(I_2\)[/tex] and find [tex]\(\omega_2\)[/tex] after the people move to the center.

4. **Change in Angular Speed:**

  Find the change in angular speed, [tex]\(\Delta \omega = \omega_2 - \omega_1\)[/tex].

5. **Apparent Acceleration:**

  Use [tex]\(a = r \times \alpha\)[/tex] and the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim [tex](\(r = 151 \, \text{m}\))[/tex] to find the new apparent acceleration [tex]\(a_2\)[/tex] caused by the change in angular speed.

I'll solve for the new apparent acceleration [tex]\(a_2\)[/tex] once the change in angular speed is calculated.

Upon calculating the initial and final angular velocities from the conservation of angular momentum, the change in angular speed [tex]\(\Delta \omega\)[/tex] is determined to be [tex]\(1.96 \times 10^{-2} \, \text{rad/s}\)[/tex].

Given the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim of the space station with [tex]\(r = 151 \, \text{m}\)[/tex], the new apparent acceleration [tex]\(a_2\)[/tex] due to the change in angular speed can be found using the relationship [tex]\(a = r \times \alpha\)[/tex].

[tex]\(a_2 = r \times \Delta \omega = 151 \, \text{m} \times 1.96 \times 10^{-2} \, \text{rad/s}\)[/tex]

This calculation yields the new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].

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Answer:

[tex]2.5 rad/s^2, 688^{\circ}[/tex]

Explanation:

The angular acceleration is given by:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 74.0 rev/min \cdot (\frac{2\pi rad/rev}{60 s/min})=7.75 rad/s[/tex] is the final angular speed

[tex]\omega_i = 0[/tex] is the initial angular speed

t = 3.10 s is the time interval

Solving the equation,

[tex]\alpha = \frac{7.75 rad/s - 0}{3.10 s}=2.5 rad/s^2[/tex]

Now we can find the angular displacement by using:

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

Substituting,

[tex]\theta=0+\frac{1}{2}(2.5 rad/s^2)(3.10 s)^2=12.0 rad[/tex]

In degrees:

[tex]\theta = \frac{12.0 rad}{2\pi}\cdot 360^{\circ}=688^{\circ}[/tex]

Answer:

Torque is defined as the force that produces rotation around the axis of an object. Torque is also known as the moment of force. It's S.I unit is Nm.

Formula of torque is:

                τ= rxF(sinθ)

Given Data:

Force = 20 N

r = 0.25 m

θ = 120°

Solution:

              τ= rxF(sinθ)

              = 0.25 x 20 (sin 120)

              =  2.903 Nm

Thus, the total torque applied on the wrench is 2.903 Nm.

The magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.

To determine the magnitude of the torque when a force is applied at an angle to a wrench, we can use the formula:

Torque = Force x Distance x sin(angle)

In this case, the force is 20 N, the distance is 0.25 meters, and the angle is 120 degrees. Plugging in these values, we get:

Torque = 20 N x 0.25 m x sin(120°)

Torque = 20 N x 0.25 m x 0.866

Torque = 4.33 N.m

Therefore, the magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.

Answer:

2.23 s

Explanation:

For a second-order reaction:

1 / [A] = 1 / [A]₀ + kt

Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:

1 / 0.330 = 1 / 0.750 + 0.760t

t = 2.23

It would take 2.23 seconds.

Final answer:

To find the time it takes for a second-order reaction concentration to decrease from 0.750 M to 0.330 M at the given rate constant, we use the integrated second-order rate law. After substituting the given values and solving for time, we find that it takes approximately 1.20 seconds.

Explanation:

To calculate the time taken for the concentration of A to decrease from 0.750 M to 0.330 M in this second-order reaction, we can use the integrated second-order rate law which is given by:

[tex]\( \frac{1}{[A]_t} - \frac{1}{[A]_0} = k \cdot t \)[/tex]

Where [tex]\([A]_t\)[/tex] is the concentration of A at time t, [tex]\([A]_0\)[/tex] is the initial concentration of A, k is the rate constant, and t is the time.

Plugging in the values, we get:

[tex]\( \frac{1}{0.330 \ M} - \frac{1}{0.750 \ M} = 0.760 \ M^{-1}s^{-1} \cdot t \)[/tex]

After solving for t, we find that the time required is approximately 1.20 seconds. Note that the units of k given [tex](\( M^{-1}s^{-1} \))[/tex] are consistent with a second-order reaction, which confirms the correctness of the approach.

Answer:

Electric field, E = 2 N/C

Explanation:

It is given that,

Mass of an electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]

Electric force, [tex]F=3.2\times 10^{-19}\ N[/tex] (in +x direction)

We have to find the electric field at this point. The electric field is given by the electric force acting on a charged particle per unit electric charge. Mathematically, it is given by :

[tex]E=\dfrac{F}{q}[/tex]

q = charge on an electron, [tex]1.6\times 10^{-19}\ C[/tex]

[tex]E=\dfrac{3.2\times 10^{-19}\ N}{1.6\times 10^{-19}\ C}[/tex]

E = 2 N/C

The electric field at this point is 2 N/C.

Final answer:

To find the weight of the tool on Newtonia, calculate the acceleration using the given displacement and time. Multiply this acceleration by the mass of the tool to find its weight. On Earth, use the acceleration due to gravity to find the weight of the tool.

Explanation:

To determine the weight of the tool on Newtonia, we can use the equation for weight: w = mg. From the given information, we know that the tool moves 16.1 m in the first 2.40 s when a 12.2 N force is applied. Using this information, we can calculate the acceleration of the tool. By using the equation of motion s = ut + 0.5at^2, where s is the displacement, u is the initial velocity (0 m/s), t is the time (2.40 s), and a is the acceleration, we can solve for a. The calculated acceleration can then be multiplied by the mass of the tool to find its weight on Newtonia.

To find the weight of the tool on Earth, we can use the acceleration due to gravity, which is approximately 9.8 m/s². Again using the equation w = mg, we can multiply the mass of the tool by this acceleration to find its weight on Earth.

Answer:

Option B is the correct answer.

Explanation:

Shear stress is the ratio of shear force to area.

We have

       Shear stress = 3 N/mm² = 3 x 10⁶ N/m²

       Area = Area of rectangle = 10 x 10⁻² x d = 0.1d

       Shear force = 50000 N

Substituting

        [tex]\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm[/tex]

Width of beam = 16.67 cm

Option B is the correct answer.

To calculate the depth of the beam, shear stress is divided by shear force to get the cross-sectional area, which is then divided by the width to obtain the depth. The depth is found to be 16.67 cm.

The subject of the question relates to the calculation of the depth of a rectangular beam based on the given maximum shear force and shear stress. The formula to be used involves shear stress (τ), shear force (V), and the cross-sectional area (A), which is the product of the beam's width and depth (b × d).

Calculation:

Shear Stress (τ) = Shear Force (V) / Area (A)

Given that the maximum shear stress is 3 N/mm² and the shear force is 50000 N, we can rearrange the formula to solve for the cross-sectional area (A):

A = V / τ = 50000 N / 3 N/mm²

Convert 50000 N to mm² by multiplying by 1000 to maintain the correct units:

A = 50000 N × 1000 mm²/N = 50000000 mm²

A = 50000000 mm² / 3 N/mm² = 16666666.67 mm²

Next, we find the depth, remembering the width of the beam is 10 cm (which is 100 mm because 1 cm = 10 mm):

Depth (d) = A / width (b) = 16666666.67 mm² / 100 mm

Depth (d) = 166666.67 mm which is equal to 16.67 cm.

Therefore, the correct answer is (c) 16.67 cm.

Final answer:

Magnetic flux is a representation of the magnetic field passing through an area—it's calculated using the field strength, area vector, and the cosine of their angle. Faraday's law of induction explains that changing magnetic flux through a loop induces an emf in it, which is crucial in electric generators and related technologies. Lenz's law ensures the direction of the induced current opposes the flux change, reflecting energy conservation.

Explanation:

Magnetic Flux and Faraday's Law of Induction

Magnetic flux is essentially a measure of the quantity of magnetic field lines passing through a given area. To determine it, you need to know the strength of the magnetic field, the surface area over which it's spread, and the angle between the field and the normal (perpendicular) to that surface area. Mathematically, it is defined as the product of the magnetic field (B), the area (A), and the cosine of the angle (θ) between them, which is given by the equation Φ = B * A * cos(θ).

Faraday's law of induction tells us that a changing magnetic flux within a closed conducting loop will induce an electromotive force (emf) in the loop. This phenomenon is used in a vast array of technologies from electric generators to transformers. The magnitude of the induced emf can be calculated using Faraday's law which states that the emf induced is directly proportional to the rate of change of the magnetic flux.

Additionally, Lenz's law elaborates on the direction of the induced current, stating that it will flow in a manner that opposes the change in flux that caused it. This is a manifestation of the conservation of energy.

Application Example

An example of Faraday's law would be a coil of wire exposed to a changing magnetic field. If the field changes over time, an electric current is induced in the wire. This principle is the foundation for devices like inductive chargers and electrical generators.

Answer:

The speed of the plank is 81.68 m/s

Explanation:

Given that,

Speed of bullet = 152 m/s

Speed of wood = 128 m/s

Speed of another bullet = 97 m/s

We need to calculate the speed of plank

Using conservation of momentum

[tex]m_{1}u_{1}=(m_{1}+m_{2})v[/tex]

Where,

u = initial velocity

v = final velocity

[tex]152m_{1}=(m_{1}+m_{2})128[/tex]....(I)

[tex]m_{1}97=(m_{1}+m_{2})v[/tex]....(II)

From equation(I) and equation(II)

[tex]\dfrac{152}{97}=\dfrac{128}{v}[/tex]

[tex]= \dfrac{128\times97}{152}[/tex]

[tex]v=81.68\ m/s[/tex]

Hence, The speed of the plank is 81.68 m/s

Answer:

B. Remain the same

Explanation:

Answer:

p = 20, q = 50

Explanation:

p² + 16q = 1200

300 - p² + 2q = 0

By adding the two equations, we can eliminate p²:

16q + 300 + 2q = 1200

18q = 900

q = 50

Solving for p:

p² + 16(50) = 1200

p = 20

Final answer:

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars, after solving the system of equations set by equating the demand function p^2 + 16q = 1200 and supply function 300 - p^2 + 2q = 0.

Explanation:

To find the equilibrium quantity and equilibrium price for the given demand function p2 + 16q = 1200 and supply function 300 − p2 + 2q = 0, we need to set the quantity demanded equal to the quantity supplied and solve the system of equations.

Firstly, we rearrange the supply function to isolate p2, which yields p2 = 300 - 2q. Next, we substitute this expression into the demand function equation:

(300 - 2q) + 16q = 1200

Now, we solve for q:

300 + 14q = 1200

14q = 900

q = 900 / 14

q = 64.29 (rounded to two decimal places)

Then, we substitute q back into either the original demand or supply function to find p. Using the demand function:

p2 + 16(64.29) = 1200

p2 + 1028.64 = 1200

p2 = 171.36

[tex]p = \sqrt{171.36}[/tex]

p = 13.09 (rounded to two decimal places)

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars.

Answer:

1.23 J/(g °C)

Explanation:

[tex]m_{w}[/tex] = mass of water = 50 g

[tex]c_{w}[/tex] = specific heat of water = 4.186 J/(g °C)

[tex]T_{wo}[/tex] = Initial temperature of water = 22.00 °C

[tex]m_{a}[/tex] = mass of alloy = 25 g

[tex]c_{a}[/tex] = specific heat of alloy = ?

[tex]T_{ao}[/tex] = Initial temperature of alloy = 93.00 °C

[tex]T_{f}[/tex] = Final equilibrium temperature = 31.10 °C

Using conservation of heat

Heat Lost by alloy = Heat gained by water

[tex]m_{a}c_{a}(T_{ao}-T_{f})[/tex] = [tex]m_{a}c_{a}(T_{f} - T_{ao})[/tex]

(25) [tex]c_{a}[/tex] (93 - 31.10) = (50) (4.186) (31.10 - 22)

[tex]c_{a}[/tex] = 1.23 J/(g °C)

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

[tex]\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}[/tex]

Where, [tex]\beta=\dfrac{c}{v}[/tex]

[tex]\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}[/tex]

[tex]\lambda_{o}=115\ nm[/tex]

Hence, The observer detects light of wavelength is 115 nm.

Answer:

Beats is a phenomenon which occurs when two sound waves with different frequencies overlap each other.

Explanation:

When these two sound waves overlap, a third frequency is formed. The places where the two waves reinforces each other, constructive interference takes place and a louder sound is produced whereas, at points where destructive interference takes place a softer sound is produced.

These alternative loud and soft sounds are called beats.

Answer:

Speed of particle, v = 51.6 m/s

Explanation:

It is given that,

Initial speed of the particle which is moving towards east, u = 20 m/s

Acceleration of the particle, a = 3.95 m/s²

Time taken, t = 8 s

We have to find the speed of the particle after this acceleration. Let the speed is v. It can be calculated using first equation of motion as :

v = u + at

[tex]v=20\ m/s+3.95\ m/s^2\times 8\ s[/tex]

v = 51.6 m/s

So, the speed of the particle after this acceleration is 51.6 m/s

Answer:

[tex]4.41\cdot 10^5 N[/tex]

Explanation:

First of all, let's calculate the total mass of the train+the engines:

[tex]m=2(8.00\cdot 10^4 kg) + 45(5.50\cdot 10^4 kg) = 2.64\cdot 10^6 kg[/tex]

Then we can apply Newton's second law, which states that the resultant of the forces is equal to the product between mass (m) and acceleration (a):

[tex]\sum F = ma[/tex] (1)

In this case there are two forces:

- The pushing force exerted by the engines, F

- The frictional force, [tex]F_f = 7.50 \cdot 10^5 N[/tex], in an opposite direction to the acceleration

So (1) becomes

[tex]F-F_f = ma[/tex]

Since the acceleration must be

[tex]a=5.00\cdot 10^{-2} m/s^2[/tex]

We  can solve the formula to find F:

[tex]F=ma+F_f = (2.64\cdot 10^6 kg)(5.00\cdot 10^{-2} m/s^2) + 7.50 \cdot 10^5 N = 8.82\cdot 10^5 N[/tex]

However, this is the force exerted by both engines. So the force exerted by each engine must be half this value:

[tex]F=\frac{8.82\cdot 10^5 N}{2}=4.41\cdot 10^5 N[/tex]

The person exerts a force of 75 N on the mower, and it will travel approximately 0.529 meters before coming to a stop.

To find the force (F) exerted by the person on the mower, we can use Newton's second law, which states:

[tex]\[ \text{Net force} = \text{Mass} \times \text{Acceleration} \][/tex]

Given:

- Mass of the mower (m) = 24 kg

- Net external force (F_net) = 51 N

- Force of friction (F_friction) = 24 N

We know that net external force is the sum of all forces acting on the object, so:

[tex]\[ F_{\text{net}} = F - F_{\text{friction}} \][/tex]

[tex]\[ 51 \, \text{N} = F - 24 \, \text{N} \][/tex]

[tex]\[ F = 51 \, \text{N} + 24 \, \text{N} \][/tex]

[tex]\[ F = 75 \, \text{N} \][/tex]

So, the person exerts a force of [tex]\( 75 \, \text{N} \)[/tex] on the mower.

Now, to find how far the mower will go before stopping, we can use the equation of motion:

[tex]\[ v^2 = u^2 + 2as \][/tex]

Where:

- \( v \) = final velocity (0 m/s, since the mower stops)

- \( u \) = initial velocity = 1.5 m/s

- \( a \) = acceleration

- \( s \) = distance traveled

We know that the net force acting on the mower is responsible for the acceleration, and it's given by Newton's second law:

[tex]\[ F_{\text{net}} = m \times a \][/tex]

So:

[tex]\[ a = \frac{F_{\text{net}}}{m} \][/tex]

[tex]\[ a = \frac{51 \, \text{N}}{24 \, \text{kg}} \][/tex]

[tex]\[ a ≈ 2.125 \, \text{m/s}^2 \][/tex]

Now, using the equation of motion:

[tex]\[ 0^2 = (1.5 \, \text{m/s})^2 + 2 \times 2.125 \, \text{m/s}^2 \times s \][/tex]

[tex]\[ 0 = 2.25 + 4.25s \][/tex]

[tex]\[ 4.25s = -2.25 \][/tex]

[tex]\[ s = \frac{-2.25}{4.25} \][/tex]

[tex]\[ s ≈ -0.529 \, \text{m} \][/tex]

The negative sign indicates that the mower moves in the opposite direction to its initial motion before stopping. So, the mower will travel approximately [tex]\( 0.529 \, \text{m} \)[/tex] before coming to a stop.

The force exerted by the person on the mower is 75 N. The mower will travel approximately 1.125 meters before stopping.

The net external force is the difference between the force exerted by the person and the force of friction. Thus, we can write:

[tex]F_{net} = F - F_{friction}[/tex]

Substituting the given values:

[tex]51\,\text{N} = F - 24\,\text{N}[/tex]

Solving for F:

[tex]F = 51\,\text{N} + 24\,\text{N} = 75\,\text{N}[/tex]

Finding the Distance the Mower Travels before Stopping:

When the force F is removed, the mower will decelerate due to the frictional force. First, we find the acceleration (a) using Newton's second law:

[tex]F_{friction} = m \cdot a[/tex]

Solving for a gives:

[tex]a = \frac{F_{friction}}{m} = \frac{24\,\text{N}}{24\,\text{kg}} = 1\, \text{m/s}^2[/tex]

Since friction opposes the motion, this is a deceleration (a is negative):

[tex]a = -1\, \text{m/s}^2[/tex]

Next, we use the kinematic equation to find the distance (d):

[tex]v^2 = v_0^2 + 2 \cdot a \cdot d[/tex]

Substituting the known values:

[tex]0 = (1.5 \frac{m}{s})^2 + 2 \cdot (-1 \frac{m}{s^2}) \cdot d[/tex]

Solving for d:

[tex]0 = 2.25 - 2d[/tex]
[tex]2d = 2.25[/tex]
[tex]d = 1.125 \text{m}[/tex]

Answer:

20 bowling balls

Explanation:

Answer:

20 bowling balls

Explanation:

A hot air balloon would have to have little mass, actually less mass than the air surrounding it in order to be able to float, a canoe ussualy has a mass of around 100 lb, and 30 ounces of lead obviusly are not more than 20 bowling balls, the average mass of a bowling mass is 14 lb, if you multiply that by 20 that is 280 pounds of mass, so the 20 bowling balls have the most mass.

Answer:

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Explanation:

As we know that flux of magnetic field from the closed loop is given by

[tex]\phi = BAcos30[tex]

now by Faraday's law we know

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we will have

[tex]EMF = Acos30\frac{dB}{dt}[/tex]

now we have

[tex]A = 1 m^2[/tex]

EMF = 10 Volts

[tex]10 = 1 (cos30)\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt} = \frac{20}{\sqrt3}[/tex]

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

g = acceleration due to gravity on moon

[tex]g=\dfrac{4\pi^2l}{T^2}[/tex]

[tex]g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}[/tex]

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

If the pendulum of length is 1.0 m is pulled to the side and released on the moon and It's period is measured to be 4.82 seconds, the gravity on the moon is 1.69 m/s².

Explanation:

It is given that,

Mass of the car, m = 1200 kg

Radius of circular path, r = 84 m

The coefficient of static friction between tires and road is 0.60

Static friction is given by :

[tex]f=\mu N=\mu mg[/tex]..............(1)

When an object is moving in a circular path, the centripetal force will act on it. It is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

Let v is the maximum speed of the car. From equation (1) and (2) as :

[tex]v=\sqrt{\mu gr}[/tex]

g = acceleration due to gravity

[tex]v=\sqrt{0.6\times 9.8\ m/s^2\times 84\ m}[/tex]

v = 22.22 m/s

or

v = 22 m/s

Hence, this is the required solution.

Answer:

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) The maximum height attained by the rocket from the ground is 348.65 m

(c) The time taken by the rocket to maximum height after lift off is 8.85 s.

(d) The total time taken by the rocket in air is 17.3 second.

Explanation:

u = 57 m/s, a = 3 m/s^2, h = 140 m

let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.

Use III equation of motion

V^2 = u^2 + 2a h

v^2 = 57^2 + 2 x 3 x 140

v = 63.95 m/s

Now use I equation of motion

v = u + at

t = (63.95 - 57) / 3 = 2.32 s

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) Let H be the maximum height reached by the rocket after the engine stops.

Use III equation of motion

v^2 = u^2 + 2aH

here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2

0 = 63.95^2 - 2 x 9.8 x H

H = 208.65 m

The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m

(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.

Use I equation of motion

v = u + a t'

0 = 63.95 - 9.8 x t'

t' = 6.53 s

The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.

(d) let t'' be the time taken by the rocket to fall freely

Use II equation of motion

H' = ut'' + 1/2 gt''^2

Here, H' = 348.65 m, u = 0

348.65 = 0 + 0.5 x 9.8 x t''^2

t''^ = 8.44 s

The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.

Answer:

Mass , momentum and mechanical energy

Explanation:

There are two types of collision.

1. Elastic collision: in case of elastic collision no energy is loss in form of heat and sound .

The momentum and the kinetic energy is conserved.

2. Inelastic collision: in case of inelastic or partially elastic collision, some of energy is lost in form of heat or sound. Here momentum and mechanical energy is conserved.

Mass of a system is always conserved.

Answer:

0.28 cm

Explanation:

The volume of a sphere is given by:

[tex]V=\frac{4}{3}\pi r^3[/tex]

where r is the radius, which is dependent on the time, so r(t).

The rate of change of the volume is

[tex]\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}[/tex] (1)

where

[tex]\frac{dr}{dt}[/tex] is the rate of change of the radius. We know that

[tex]\frac{dr}{dt}=9[/tex] (cm/s)

And we want to find the value of the radius r when the rate of change of the volume is the same:

[tex]\frac{dV}{dt}=9[/tex] (cm^3/s)

So we can rewrite (1) as:

[tex]9=4\pi r^2 \cdot 9[/tex]

By solving it, we find

[tex]4\pi r^2 = 1\\r = \sqrt{\frac{1}{4\pi}}=0.28 cm[/tex]

Answer:

g

Explanation:

gg

a.Breccia contains angular clasts and conglomerate contains rounded clasts.

Breccia and conglomerate are rocks that can primarily be distinguished by the shape of their clasts; breccia has angular fragments while the fragments in conglomerate are rounded.

One can distinguish between breccia and conglomerate based on the nature of their clasts. Breccia contains angular clasts, meaning its fragments have not been smoothed or rounded and retain sharp, rough edges. On the other hand, conglomerate contains rounded clasts.

The fragments within a conglomerate have been weathered and eroded over time, resulting in more rounded and smooth shapes. Both of these sedimentary rocks are formed from fragments of igneous rock or the shells of living organisms, but their difference lies in the shape and wear of their components.

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Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

[tex]\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s[/tex]

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

[tex]a=\omega^2 r[/tex]

At the instant t = 0,

[tex]\omega=3.93 rad/s[/tex]

So the centripetal acceleration of the tip of the blades is

[tex]a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2[/tex]