QUESTION 2

Objectives:

I . identify potential and kinetic energy in a situation and draw corresponding energy bar charts,

II . calculate gravitational and elastic potential energy,

III . draw & analyze potential energy functions, and (d) use conservation of energy to relate the total energy at one time to total energy at another time.

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Bungee Jump: you will step off with zero initial vertical velocity from a platform a height h above the ground. The bungee cord will act like a giant extensional spring that will, you hope, provide an upward force on becoming taut. After weighing you (you have a mass M), the operator has selected a bungee cordwith an un-stretched length of d and a spring constant of k.

Consider yourself to be a single point – i.e., use the particle model.

Choosing the ground as your origin (and the z-axis directed upwards), answer the following questions about your bungee jumping adventure in terms of M, h, d, k, z, and the gravitational field strength, g. Answer with variables.

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a)

Write an expression for the stretching ?L of the cord in terms of d and z and for the total potential energy U of the jumper-bungee-Earth system for each situation (consider the latter two situations together).

b)

Which type of potential energy ( UG or US ) is largest for large z (early in the fall)? For small z (late in the fall)?

c)

Sketch a graph of your gravitational potential energy, UG(z) vs your height, z, from z = 0 to z = h, on the left plot. Then sketch a graph of your elastic potential energy, US(z) on the center plot. Finally on the rightmost plot, sketch a graph of your total potential energy1 U(z) = UG(z) + US(z). Do these plots on your own without the help of a computer or calculator.

Answer:

B)

Explanation:

Hi there,

Notice in the prompt it is saying given mass, which means quantity of gas is constant. This eliminates option C, as this is Avogadro's Law (the more gas you have, the more space it takes up). It also states pressure remains constant, so it eliminates D, Boyle's Law (pressure is inversely related with volume), and A, The Combined Gas Law (combines Charles, Boyle's, and Gay-Lussac's Laws).

Charles' Law is telling us that as a gas is heated, the volume increases. This makes sense, as when gases are heated the average kinetic energy (energy of motion and movement) of the system is higher because of heat increase. When the system's kinetic energy is higher, it will have the tendency to occupy more space to balance out the increase in energy. Think of people who become more jittery in an already crowded space, they will push outward to accompany more space to balance.

So, answer is B.

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thanks,

Answer:2.7m/s^2

Explanation:

mass=2.3kg

Force=6.2Newton

Acceleration=force ➗ mass

Acceleration=6.2 ➗ 2.3

Acceleration=2.7m/s^2

The resulting acceleration of the object is  2.70 m/s^2.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

Given that:

Mass of the object: m = 2.3 kg.

Force applied on it: F = 6.2 Newtons.

Now from Newton's 2nd law of motion: it can be stated that:

force = mass × acceleration

acceleration = force/mass

= 6.2 Newton/2.3 kg

= 2.70 m/s^2.

Hence,  the acceleration of the object is 2.70 m/s^2.

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Answer:

p = 4/3   f

Explanation:

The constructor station describes the position of the image object in relation to the focal length

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distances to the object and the image respectively

 the objent distance  is

         1 / p = 1 / f - 1 / q

they indicate that the image is located at q = 4f this distance is positive because the image is after the lens

        1 / p = 1 / f - 1 / 4f

        1 / p = 3 / 4f

         p = 4f / 3

the object is positive because it is to the left of the lens, according to the optical sign convention

The sign convention is that the distance to the objects is positive if it is on the left side of the lens and the image is positive if it is on the right side of the lens.

Newton is used to measure force

The most used unit for the force measurement is "newton" which is the SI unit of the force

Answer:

newtons

Explanation:

Answer:

68.46 m.

Explanation:

Given,

coefficient of friction,μ = 0.67

speed of truck, v = 30 m/s

distance travel by the truck to stop = ?

Now,

Calculation of acceleration

we know,

f = m a

and also

f = μ N = μ mg

Equating both the forces equation

m a = μ mg

a = μ g

a = 0.67 x 9.81

a = 6.57 m/s²

Now, using equation of kinematics

v² = u² + 2 a s

0 = 30² - 2 x 6.57 x s

s = 68.46 m

Hence, the minimum distance travel by the truck is equal to 68.46 m.

The truck would need to stop in a minimum distance of approximately 68.5 meters to prevent the box from sliding off, based on the given parameters and considering the law of inertia and the role of static friction.

To determine the minimum distance the truck can stop without the box sliding off, we need to consider the law of inertia and the role of static friction.

The force must meet or exceed the force exerted by the truck's deceleration to keep the box from sliding.

Given that the initial velocity of the truck is 30 m/s and the box has a mass of 500 kg, we can use the formula for static friction, which is F = μN, where F is the force of friction, μ is the coefficient of static friction, and N is the normal force.

In this case, N equates to the gravitational force on the box, so N = mg = 500 kg * 9.8 m/s² = 4900 N. Substituting these values in the formula for static friction, we get F = 0.67 * 4900 N = 3283 N.

The force of friction, in terms of motion, is also equivalent to mass times acceleration (F = ma), so we can set this equal to the static friction we calculated and solve for acceleration: 3283 N = 500 kg * a. Solving for a gives an acceleration of approximately 6.57 m/s².

Using motion equations, specifically v² = u² + 2as where v is the final velocity (0 m/s), u is the initial velocity (30 m/s), a is the acceleration (-6.57 m/s² because it's deceleration), and s is the distance, we can compute for the distance s which gives a value of approximately 68.5 meters.

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The probable question is :-

A truck is moving along a straight horizontal road at 30 m/s. A box is placed on the bed of the truck, and the coefficient of static friction between the bed and the box is 0.67. What is the least distance in which the truck can come to a stop without causing the box to slide? Consider the deceleration of the truck due to braking.

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

[tex]\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s[/tex]

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

[tex]\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s[/tex]

So the flow speed of each of the narrower pipe is:

[tex]v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}[/tex]

[tex]v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s[/tex]

Answer: 140cm

Explanation:

Radius of curvature r = 2f

Where f is the focal point

f = 70cm, therefore

r =2 x 70 = 140cm

Answer:

The radius of curvature is 9.35 cm.

Explanation:

Given:

u = -5 cm

v = -70 cm

The radius of curvature of the mirror can be obtained from the expression:

[tex]\frac{1}{v} +\frac{1}{u} =\frac{2}{R} \\-\frac{1}{v} -\frac{1}{u} =\frac{2}{R} \\-\frac{1}{70} -\frac{1}{5} =\frac{2}{R} \\-0.014-0.2=\frac{2}{R} \\R=-9.35cm[/tex]

Answer:

c.[tex]a_m[/tex]

Explanation:

We are given that

Acceleration due to gravity on the moon=[tex]a_m[/tex]

Acceleration due to gravity on the earth=[tex]a_e[/tex]

[tex]g_m=\frac{1}{6}g_e[/tex]

Net force due to am on an object on moon=[tex]F_{net}=ma_m[/tex]

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=[tex]F=ma_e[/tex]

[tex]F=F_{net}[/tex](given)

[tex]ma_m=ma_e[/tex]

[tex]a_e=a_m[/tex]

Hence, option c is true.

Explanation:

We have,

Initial speed of an automobile, u = 71 km/h = 19.72 m/s

Diameter of the tie, d = 60 cm

Radius, r = 30 cm

(a) The angular speed of the tires about their axles is given by :

[tex]\omega=\dfrac{v}{r}\\\\\omega=\dfrac{19.72}{0.3}\\\\\omega=65.73\ rad/s[/tex]

(b) Final angular velocity of the wheel is equal to 0 as its stops. The angular acceleration of the wheel is given by :

[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]

[tex]\theta=40\ rev\\\\\theta=251.32\ rad[/tex]

[tex]0-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_i^2}{2\theta}\\\\\alpha =\dfrac{(65.73)^2}{2\times 251.32}\\\\\alpha =-8.59\ rad/s^2[/tex]

(c) Let the car move a distance d during the braking. So,

[tex]d=\theta r\\\\d=251.32\times 0.3\\\\d=75.39\ m[/tex]

Therefore, the above is the required explanation.

a. The angular speed is 65.73 rad/s.

b. The magnitude of the angular acceleration is -8.59 rad/s².

c. The distance should be 75.39m.

Since

Initial speed of an automobile, u = 71 km/h = 19.72 m/s

Diameter of the tie, d = 60 cm

Radius, r = 30 cm

a. Now the angular speed should be

= v/r

= 19.72/0.3

= 65.73 rad/s

b. Now the magnitude is

= 65.73^2/2*251.32

= -8.59 rad/s^2

c. The distance should be

= 251.32*0.3

= 75.39 m

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Answer:

The magnitude of the force exerted by the ball on the ground during the 0.5 s of contact = 90.16 N

Explanation:

Given,

Mass of mud ball = m = 2.40 kg

Height the ball is released from = y = 18 m

Total contact time of ball and the ground = t = 0.5 s

The Newton's second law of motion explains that the magnitude of the change of momentum is equal to the magnitude of a body's impulse.

Change in momentum = Magnitude of Impulse

Change in momentum = (final momentum) - (initial momentum)

Since the ball is dropped from rest, initial momentum = 0 kgm/s

But to calculate its final momentum, we need the ball's final velocity before hitting the ground.

Using the equations of motion,

u = initial velocity of the ball = 0 m/s (ball was dropped from rest)

v = final velocity of the ball = ?

g = acceleration due to gravity = 9.8 m/s²

y = vertical distance covered by the ball = 18 m

v² = u² + 2gy

v² = 0² + (2)(9.8)(18)

v² = 352.8

v = 18.78 m/s

Final momentum of the ball = (m)(v)

= (2.4) × (18.78) = 45.08 kgm/s

Change in momentum = 45.08 - 0 = 45.08 kgm/s

Impulse = Ft

Change in momentum = Magnitude of Impulse

45.08 = F × (0.5)

F = (45.08/0.5) = 90.16 N

Hope this Helps!!!

Answer:

90.14 N

Explanation:

according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formula below

v² = u² + 2gh

Where h = height of the cliff = 18.0m

v² = 2 × 9.8 × 18

v² = 352.8

v = √333.2

v = 18.78 m/s

At t = 0.50s and v = 18.78 m/s, we can get the average force of impact

F×0.50 = 2.4 (18.78 - 0)

F × 0.50 = 2.4 (18.78)

F × 0.50 = 45.072

F = 45.072 /0.50

= 90.14 N

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

[tex]W = E_p = mgh [/tex]

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

[tex]W = 50*9.8*13 = 6370 J[/tex]

Final answer:

The width of the slit is 1.0 μm.

Explanation:

When light passes through a single slit, it undergoes diffraction, which causes interference patterns. The width of the central peak in the diffraction pattern is related to the width of the slit and the wavelength of the light. In this case, the width of the central peak is given as 5.0 mm and the wavelength is given as 600 nm.

Using the formula for the width of the central peak, we can solve for the width of the slit:

Width of slit = (wavelength * distance to screen) / (number of the peak * distance to the peak)

Substituting the given values into the formula, we find that the width of the slit is 1.0 μm.

Answer: The approximate magnetic field is 1.38 × 10^-3T

Explanation: Please see the attachment below

The approximate magnetic field at a point within the solenoid is 4.4 × 10^-4 T.

To find the magnetic field strength at a point within a solenoid, we can use the equation B = µ0nI, where B is the magnetic field strength, µ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid carries a current of 10.0 A and has 110.0 turns per meter of its length, we can substitute these values into the equation to calculate the magnetic field strength:

B = (4π × 10-7 T·m/A)(110.0 turns/m)(10.0 A) = 4.4 × 10-4 T

Therefore, the approximate magnetic field at a point within the solenoid is 4.4 × 10-4 T.

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Answer:

3.53 N/C

Explanation:

Electric field = F / q where F is the force in N and q is charge on the electron

F = mass of an electron × a ( acceleration in m/s)

using equation of motion to solve for the acceleration

s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0

2s / t² = a

F = me × ( 2s / t²)

E electric field = me × ( 2s / t²)  / q = me × 2s / ( t² × q)

me, mass of an electron = 9.11 × 10⁻³¹ kg

E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C

Answer: The magnitude of the electric field is 3.53 N/C

Explanation: Please see the attachments below

Answer:

AM radio, FM radio, microwaves, sodium light.

Explanation:

Electromagnetic radiation are waves from electromagnetic field which spread through the space or any other material medium and carries radiating energy. Examples includes X rays, radio waves, Gamma rays, etc. Exposure to high level of electromagnetic radiation could be harmful to human body, on the other hand, science as not been able to prove that exposing humans to low level electromagnetic radiation is harmful to our health.

The mass of the solid cylinder is calculated by setting up a triple integral over the volume of the cylinder using cylindrical coordinates with the given density function. The integral is calculated from the inside out, starting with the integral over z (height), followed by r (radius), and finally θ (angular measure).

The mass of a solid object in three-dimensions using a cylindrical coordinate system (r,θ,z) can be expressed as an integral over the volume of the object multiplied by the density function. In your case, given that the solid cylinder D equals {(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ z ≤ 10} with density function ρ(r, θ, z) = 1+ z/2, we calculate mass M as follows:

M=∫02π∫02∫010(1+ z/2)rdzdrdθ

Here, we start by calculating the innermost integral (∫010(1+ z/2)dz) first, then move to the middle integral (∫02dr) and finally the outermost integral (∫02πdθ). Each integral works on the outcome of the next inner integral until the entire mass of the cylinder is calculated.

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The mass of the solid cylinder is [tex]\(140\pi\).[/tex]

The mass of the solid cylinder is given by the triple integral

[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]

To find the mass of the solid cylinder, we need to evaluate the triple integral of the density function over the volume of the cylinder in cylindrical coordinates. The density function is given by [tex]\(\rho(r, \theta, z) = 1 + \frac{z}{2}\).[/tex]

The solid cylinder is defined by the set [tex]\(D = \{(r, \theta, z) : 0 \leq r \leq 2, 0 \leq z \leq 10\}\)[/tex]. The limits of integration for[tex]\(r\), \(z\), and \(\theta\)[/tex]  are determined by the geometry of the cylinder:

- For[tex]\(r\)[/tex], the radial distance from the z-axis, the limits are from 0 to 2, since the radius of the cylinder is 2 units.

- For[tex]\(z\),[/tex] the height of the cylinder, the limits are from 0 to 10, since the height is 10 units.

- For [tex]\(\theta\),[/tex] the angle around the z-axis, the limits are from 0 to [tex]\(2\pi\),[/tex] since a full rotation around the z-axis is [tex]\(2\pi\)[/tex] radians.

The differential volume element in cylindrical coordinates is [tex]\(r \, dz \, dr \, d\theta\),[/tex] where the [tex]\(r\)[/tex]  factor accounts for the Jacobian of the transformation from Cartesian to cylindrical coordinates.

Now, we can set up the triple integral as follows:

[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]

To evaluate this integral, we would first integrate with respect to [tex]\(z\)[/tex] from 0 to 10, then with respect to [tex]\(r\)[/tex] from 0 to 2, and finally with respect to [tex]\(\theta\)[/tex]  from 0 to[tex]\(2\pi\)[/tex] . The integral can be computed as follows:

1. Integrate with respect to [tex]\(z\)[/tex] first:

[tex]\[ \int_{0}^{10} \left(1 + \frac{z}{2}\right) dz = \left[z + \frac{z^2}{4}\right]_{0}^{10} = 10 + \frac{100}{4} = 10 + 25 = 35. \][/tex]

2. Next, integrate with respect to[tex]\(r\):[/tex]

[tex]\[ \int_{0}^{2} 35r \, dr = \left[\frac{35r^2}{2}\right]_{0}^{2} = \frac{35 \cdot 4}{2} = 70. \][/tex]

3. Finally, integrate with respect to [tex]\(\theta\):[/tex]

[tex]\[ \int_{0}^{2\pi} 70 \, d\theta = \left[70\theta\right]_{0}^{2\pi} = 70 \cdot 2\pi = 140\pi. \][/tex]

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

[tex]F=\frac {m(v-u)}{t}[/tex]

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

[tex]F=\frac {0.3(2--5)}{0.0015}=1400N[/tex]

The average force exerted on the ball by the floor is  1400 N.

Force is the product of mass and acceleration. The S.I unit of force is Newton (N)

To calculate the average magnitude of the force exerted on the ball by the floor, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the average force exerted on the ball by the floor is  1400 N.

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Answer:

The expression for resistance is    [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

Explanation:

Generally flow of charge at that point is mathematically given as

              [tex]J = \frac{I}{2 \pi r L}[/tex]

Where L is length of the cylinder as given the question

The potential difference that is between the cylinders is  

                 [tex]\delta V = -E dr[/tex]

 Where is the radius

Where E is the electric field that would be experienced at that point which is mathematically represented as

               [tex]E = \rho J[/tex]

Where is the [tex]\rho[/tex] is the resistivity as given the question

considering the formula for potential difference we have

                [tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]

To get V we integrate both sides

           [tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]

                [tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

According to Ohm law

           [tex]V= IR[/tex]

Now making R the subject we have

          [tex]R = \frac{V}{I}[/tex]

               Substituting for V

         [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]

The resistance of a coaxial cylindrical configuration with radii R1 and R2, length L, and resistivity rho, is given by the formula R = (rho / (2 * pi * ln(R2/R1))).

The question asks for an expression for the resistance of a coaxial cylindrical configuration, where the space between two metal cylinders of length L and radii R1 and R2 is filled with a material of resistivity rho. To find this, we utilize the formula for the resistance R of a material, which is R = rho * (L/A), where A is the cross-sectional area. However, since the current flows radially through the material between the cylinders, we need to consider the formula for resistance in terms of the radii and length of the cylinders. The resistance can be expressed as R = (rho * L) / (2 * pi * L * ln(R2/R1)), simplifying to R = (rho / (2 * pi * ln(R2/R1))).

Explanation:

answer and explanation is in the picture

Answer:

4000 N

Explanation:

First we calculate the acceleration

2as = vf^2 - vi^2

2xax0.1 = 0^2-40^2

0.2 x a = -1600

a = - 8000 m/s^2

F= ma

F= 0.5 x 8000

F 4000N

Answer:

0.358 kg

Explanation:

From simple harmonic motion,

T = 2π√(m/k)................ Equation 1

Where T = period of the spring, k = spring constant of the spring, m = mass suspended, π = pie

make m the subject of the equation

m = kT²/4π².................. Equation 2

Given: k = 15 N/m, T = 0.97 s, π = 3.14

Substitute into equation 2

m = 15(0.97²)/(4×3.14²)

m = 14.1135/39.4384

m = 0.358 kg.

Hence mass suspended = 0.358 kg

0.356kg

The period, T, of a mass on a spring of spring constant, k, in simple harmonic motion can be calculated as follows;

T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]         --------------------(i)

Where;

m = mass

From the question;

T = 0.97s

k = 15N/m

Taking π = 3.142 and substituting the values of T and k into equation (i) as follows;

0.97 = 2 x 3.142 x [tex]\sqrt{\frac{m}{15} }[/tex]

0.97 = 6.284 x [tex]\sqrt{\frac{m}{15} }[/tex]

[tex]\frac{0.97}{6.284}[/tex] = [tex]\sqrt{\frac{m}{15} }[/tex]

0.154 = [tex]\sqrt{\frac{m}{15} }[/tex]

Square both sides

0.154² = [tex]\frac{m}{15}[/tex]

0.0237 = [tex]\frac{m}{15}[/tex]

m = 0.356

Therefore, the mass that could be suspended on the spring to have an oscillation period of 0.97s when in SHM is 0.356kg

Answer:

Explanation:

loss of energy while passing through target by bullet

= 1/2 mu² - 1/2 mv² , m is mass of bullet , u is initial velocity and v is final velocity .

= 1/2 x m ( u² - v² )

= .5 x .023 x ( 1100² - 950² )

= 3536.25 J.

This loss is due to negative work done by  friction force

If friction force be F

Work done by friction force = F x .33

F x .33 = loss of kinetic energy

F x .33 = 3536.25

F = 10716 N

impulse of F

F X t , time period during which this force remains active

10716 x t = change in momentum of bullet

= .023 ( 1100 - 950 )

= 3.45

t = 3.45 / 10716

= 3.22 x 10⁻⁴ s.

Average force on the target = friction force created = 10716 N

Impulse by force on target = 10716 x 3.22 x 10⁻⁴

impulse on target = change in momentum of target

= mass of target x its velocity after impact

=  400 v

v = 10716 x 3.22 x 10⁻⁴ / 400

= 86.26 x 10⁻⁴ m /s

Answer:

The volume is 2.22x10⁵m³

Explanation:

the solution is in the attached Word file

To calculate the volume of space containing 1.0 J of electromagnetic energy near the Earth, use the formula Volume = Energy / Intensity after converting the intensity from kW/m² to W/m².

The volume of space that contains 1.0 J of electromagnetic energy can be found using the formula:

Volume = Energy / Intensity

Given that the intensity is 1.35 kW/m², you need to convert it to W/m² (1 kW = 1000 W).

After converting the intensity, you can then calculate the volume using the given energy of 1.0 J.

The volume of space near Earth containing 1.0 J of electromagnetic energy can be found by first calculating the energy density using the given intensity and speed of light, and then dividing the amount of energy by energy density, resulting in a volume of 2.22  imes 10^5 m^3.

Finding the Volume of Space Containing 1.0 J of Electromagnetic Energy

To find the volume of space near Earth that contains 1.0 J of electromagnetic energy given the intensity (I) from the sun as 1.35 kW/m2, we can use the formula for energy density (u), which is given by the equation u = I/c, where c is the speed of light. Now, to find the volume (V) that contains energy (E), we use the equation V = E/u.

First, let's convert the intensity to watts per square meter: 1.35 kW/m2 = 1350 W/m2. Next, calculate the energy density (u):
u = 1350 W/m2 / (3.0 * 108 m/s) = 4.5 * 10-6 J/m3.

With the energy density known, we can calculate the volume (V) that contains 1.0 J of energy:
V = 1.0 J / (4.5 * 10-6 J/m3) = 2.22 * 105 m3.

Therefore, a volume of 2.22 * 105 m3 near Earth contains 1.0 J of electromagnetic energy.

The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

Explanation:

A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:

[tex]U= \frac{1}{2} LI^2[/tex]

Here L is the inductance and I is the current.

Here, energy stored (U) is proportional to the number of turns (N) and the current (I).

[tex]L = \frac{\mu_0 N^2 *A}{l}[/tex]

mu not - permeability of core material

A -area of cross section

l - length

N - no. of turns in solenoid inductor

Now,given that the proportion always remains same:

[tex]\frac{N_2}{N_1} = \frac{I_1}{I_2}[/tex]

In this way the expression

[tex]\frac{1}{5} = \frac{I_1}{I_2}[/tex]

[tex]I_2 = I_1 \times 5[/tex]

Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

Answer:

Michelson

Explanation:

Answer:

Answer

Explanation:

The followings are the reasons for the reduction in  the death toll from Tornadoes occurence in the 2000s compared to 1950s;

(a) Improvement in technology and rapid natural occurence prediction. Through the use of advanced radar systems the early detection of natural occurence e.g tornadoes, providing  warning signal to the residents and tracking their possible pathways is made possible.

Fast spread of information through television channels, Radio, Social media etc., help in fast spreading of information, thereby sending early signals to people thus making them prepare for its occurence.

Improvement of protective measures, provision of housing infrastructure for the displaced. With increase in technology, the protective measures are also improving.

Quick Government and medical interventions for casualties.  Conducting Tornado drills help people to practice to take cover in a specified location during tornadoes.

TV meteorologist are able to inform viewers of the approaching tornadoes and hurricanes  which can be tracked from the time of its origin i.e., from a depression to a storm.  Advanced radar data help in delineating the possible pathways of the hurricane.

Tornadoes on the other hand form in a thunderstorm or during a frontal activity. In thunderstorms there are vortices formed which are pre-cursors of tornadoes.. A vortice when touches the ground forms a tornado. Once a tornado is on the ground it becomes easy for radars to measure windpeeds. Hence a TV meteorologist can report the intensity of the Tornado.

Answer:

-6sin(2t)

Explanation:

fayemioluwatomisin is correct until the end.

x(0)=0

[tex]x(t)=C_1cos(2t)+C_2sin(2t)\\0=C_1cos(2(0))+C_2sin(2(0))\\C_1=0[/tex]

x'(0)=-12

[tex]x'(t)=2C_2cos(2t)\\-12=2C_2cos(2(0))\\-12=2C_2\\C_2=-6[/tex]

[tex]x(t)=-6sin(2t)[/tex]

The mass-spring system's equation of motion can be determined using Hooke's Law to find the spring constant, and applying the conditions of simple harmonic motion (SHM).

The question relates to the physical concept of a spring-mass system and its simple harmonic motion (SHM). The force exerted on a spring and the resulting stretch can determine the spring constant (k), using Hooke's Law F = kx, where F is the force applied and x is the displacement from the equilibrium position. In this case, we're given that a force of 880 newtons stretches the spring 4 meters, which allows us to calculate the spring constant k = 880 N / 4 m = 220 N/m.

Now, considering that the mass (m) attached to the spring is 55 kilograms and it's released with an upward velocity (v) of 12 m/s from the equilibrium position, the initial conditions for the SHM are established. The equation of motion for a spring-mass system in SHM is x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant. The velocity is the first derivative of the displacement function, v(t) = -Aω sin(ωt + φ), and the acceleration is the second derivative of the function.

The maximum velocity and acceleration occur when the displacement is zero and maximum respectively. To fully determine the equation of motion for this particular situation, additional information about the phase constant or initial displacement may be needed, unless it is assumed that the spring is released from the equilibrium point with an upward velocity, in which case the initial displacement would be zero, and the phase constant φ could be determined.

Answer:

Velocity between points A and B will be 0.2344 m/s

Complete Question:

A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.

Answer:

a) [tex]v_{A} = 10 m/s[/tex]

b)

[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]

Explanation:

a) Velocity of point A

The velocity of point A is a combination of the translational and the rotational velocity of the racket.

[tex]v_{A} = v_{G} + v_{r}[/tex]

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = wr[/tex]

r = 25 cm = 0.25 m

w = 20 rad/s

[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]

[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]

b) Velocity of point B

At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:

[tex]v_{G} = 5 m/s[/tex]

[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]

[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]

In the y - direction, r = 30/2 = 15 cm = 0.15 m

r = 0.15 m

[tex]w_{x} = 1 rad/s[/tex]

[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]

In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]

[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]

[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s

Answer:146.26 kN

Explanation:

Given

Outside diameter of tube [tex]d_o=54\ mm[/tex]

thickness of tube [tex]t=5\ mm[/tex]

therefore inner diameter [tex]d_i=54-2\times 5[/tex]

[tex]d_i=44\ mm[/tex]

Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]

[tex]A=\frac{\pi }{4}\times 980[/tex]

[tex]A=245\pi mm^2[/tex]

Stress developed must be less than the limited value

thus [tex]\frac{P}{A}\leq \sigma [/tex]

[tex]P\leq \sigma A[/tex]

Maximum value of [tex]P=\sigma \times A[/tex]

[tex]P=190\times 245\times \pi [/tex]

[tex]P=146.26\ kN[/tex]

Final answer:

To calculate the maximum load a stainless steel tube can support, one must first find the cross-sectional area based on its outer diameter and wall thickness. Then, using the limit of the normal stress and the area, the maximum axial load can be computed.

Explanation:

To determine the maximum load P that a stainless steel tube can support when it is used as a compression member, we need to use the formula for axial stress in a cylindrical member, which is σ = P/A, where σ is the normal stress and A is the cross-sectional area. Given that the normal stress must not exceed 190 MPa, we need to first calculate the cross-sectional area. For a tube with an outside diameter (d) of 54 mm and a wall thickness (t) of 5 mm, the internal diameter (di) would be d - 2t = 54 mm - (2 x 5 mm) = 44 mm. The cross-sectional area A can be calculated using the formula for the area of a hollow circle, A = π/4 * (d2 - di2).

A = π/4 * ((55)^2 - (44)^2)

A = 855.298 * 10^-6 m2

Once we have the area, we can calculate the maximum load by rearranging the stress formula to P = σ * A. By plugging in the values for σ and A, we can find the value of P that the member can safely support. Remember to convert diameters to meters when calculating the area in square meters for consistency with the stress units of Pascals (Pa).

P = 190 * 855.298 Pa

P = 1.62 MPa