12. A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.
How much work does the woman do in lifting the child?​

Answer:

The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.

Explanation:

Optical path length ===> 2n * t = (m + 0.5) * λ

λ = 614 nm , n = 1.33

Substitute in the parameters in the equation.

∴ 2(1.33) * t = (m + 0.5) * 614

  2.66 * t = 614m + 307

  t = (614m + 307) / 2.66 .............(1)

(a) for m = 0

    t = (614m + 307) / 2.66

    t = (614(0) + 307) / 2.66

    t = 307 / 2.66

    t = 115 nm == 0.115 um

(b) for m = 1

    t = (614(1) + 307) / 2.66

    t = (614 + 307) / 2.66

    t = 921 / 2.66

    t = 346.24 nm = 0.346 um

Answer:

The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]

The magnitude of the force at A is 832.56N

Explanation:

Here, m is the mass of the beam and l is the length of the beam.

[tex]I =\frac{1}{3} ml[/tex]

[tex]I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2[/tex]

Take the moment about point A by applying moment equilibrium equation.

[tex]\sum M_A =I \alpha[/tex]

[tex]P \sin45^0 \times 3 =I \alpha[/tex]

Here, P is the force applied to the attached cable and [tex]\alpha[/tex] is the angular acceleration.

Substitute 350 for P and 586.67kg.m² for I

[tex]350 \sin 45^0 \times3=568.67 \alpha[/tex]

[tex]\alpha =1.3056rad/s^2[/tex]

The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]

Find the acceleration along x direction

[tex]a_x = r \alpha[/tex]

Here, r is the distance from center of mass of the beam to the pin joint A.

Substitute 2 m for r and 1.3056rad/s² for [tex]\alpha[/tex]

[tex]a_x = 2\times 1.3056 = 2.6112m/s^2[/tex]

Find the acceleration along the y direction.

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

Here, ω is angular velocity.

Since beam is initially at rest,ω=0

Substitute 0 for ω

[tex]a_y = 0[/tex]

Apply force equilibrium equation along the horizontal direction.

[tex]\sum F_x =ma_x\\A_H+P \sin45^0=ma_x[/tex]

[tex]A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N[/tex]

Apply force equilibrium equation along the vertical direction.

[tex]\sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y[/tex]

[tex]A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\[/tex]

Calculate the resultant force,

[tex]F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N[/tex]

The magnitude of the force at A is 832.56N

Answer:

a) Initial angular acceleration of the beam = 1.27 rad/s²

b) [tex]F_{A} = 851.11 N[/tex]

Explanation:

[tex]tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}[/tex]

Force applied to the attached cable, P = 350 N

Mass of the beam, m = 110-kg

Mass moment of the inertia of the beam about point A = [tex]I_{A}[/tex]

Using the parallel axis theorem

[tex]I_{A} = I_{G} + m(\frac{l}{2} )^{2} \\I_{G} = \frac{ml^{2} }{12} \\I_{A} = \frac{ml^{2} }{12} + \frac{ml^{2} }{4} \\I_{A} = \frac{ml^{2} }{3}[/tex]

Moment = Force * Perpendicular distance

[tex]\sum m_{A} = I_{A} \alpha\\[/tex]

From the free body diagram drawn

[tex]\sum m_{A} = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}[/tex]

P = 350 N, l = 3+ 1 = 4 m, θ = 45°

Substitute these values into the equation above

[tex]3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}[/tex]

Linear acceleration along the x direction is given by the formula

[tex]a_{x} = r \alpha[/tex]

r = 2 m

[tex]a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}[/tex]

the linear acceleration along the y-direction is given by the formula

[tex]a_{y} = r w^{2}[/tex]

Since the beam is initially at rest, w = 0

[tex]a_{y} = 0 m/s^{2}[/tex]

General equation of motion along x - direction

[tex]F_{x} + Psin \theta = ma_{x}[/tex]

[tex]F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N[/tex]

General equation of motion along y - direction

[tex]F_{x} + Pcos \theta - mg = ma_{y}[/tex]

[tex]F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y} = 830.513 N[/tex]

Magnitude [tex]F_{A}[/tex] of the force supported by the pin at A

[tex]F_{A} = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A} = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A} = 851.11 N[/tex]

To solve this problem, apply the concepts related to the centripetal acceleration as the equivalent of gravity, and the kinematic equations of linear motion that will relate the speed, the distance traveled and the period of the body to meet the needs given in the problem. Centripetal acceleration is defined as,

[tex]a_c = \frac{v^2}{r}[/tex]

Here,

v = Tangential Velocity

r = Radius

If we rearrange the equation to get the velocity we have,

[tex]v = \sqrt{a_c r}[/tex]

But at this case the centripetal acceleration must be equal to the gravitational at the Earth, then

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

[tex]v = \sqrt{(9.8)(\frac{1000}{2})}[/tex]

[tex]v = 70m/s[/tex]

The perimeter of the cylinder would be given by,

[tex]\phi = 2\pi r[/tex]

[tex]\phi = 2\pi (500m)[/tex]

[tex]\phi = 3141.6m[/tex]

Therefore now related by kinematic equations of linear motion the speed with the distance traveled and the time we will have to

[tex]v = \frac{d}{t} \rightarrow \text{ But here } d = \phi[/tex]

[tex]v = \frac{\phi}{t} \rightarrow t = \frac{\phi}{t}[/tex]

[tex]t = \frac{3141.6}{70m/s}[/tex]

[tex]t = 44.9s[/tex]

Therefore the period will be 44.9s

Answer:

[tex]T \approx 44.88\,s[/tex]

Explanation:

"Normal" gravity is equal to 9.807 meters per squared second and cylinder must rotate at constant speed in order to simplify the equation of acceleration, which is in the radial direction. The centrifugal acceleration experimented by people allow them to be on the inside surface.

[tex]g = \omega^{2}\cdot R[/tex]

The angular speed required to provide "normal" gravity is:

[tex]\omega = \sqrt{\frac{g}{R} }[/tex]

[tex]\omega = \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{500\,m} }[/tex]

[tex]\omega \approx 0.14\,\frac{rad}{s}[/tex]

The rotation period is:

[tex]T = \frac{2\pi}{\omega}[/tex]

[tex]T = \frac{2\pi}{0.14\,\frac{rad}{s} }[/tex]

[tex]T \approx 44.88\,s[/tex]

The force between two parallel conductors carrying identical currents of one ampere over a length of one meter and separated by a distance of 5 mm is 4 x 10^-5 Newtons.

The force between two parallel conductors carrying a current is used to define the unit of current, the ampere. According to the definition, one ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly 2x10^-7 N/m on each conductor. But the wires in your question are shorter and closer together.

In this case, a simplified formula can be used which is F = (2 * k * I1 * I2 * L) / r, where k = 10^-7 N/A², I1 and I2 are the currents in the wires, L the length of the wires, and r is the distance between the wires. With I1 = I2 = 1 A, L = 1 m, and r = 5 mm = 0.005 m, the formula gives: F = (2 * 10^-7 N/A² * 1 A * 1 A * 1 m) / 0.005 m = 4 * 10^-5 N.

Therefore, the force between these two wires is 4 * 10^-5 Newtons.

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The force between the two sections of wire carrying a current of 1.0 A each and separated by a distance of 5.0 mm is 1x10*-9 N per meter

The force between two 1.0-meter-long sections of very long wires carrying a current of 1.0 A each can be calculated using the definition of the ampere. According to the definition, one ampere of current through each of two parallel conductors separated by one meter in empty space causes a force of exactly 2x10-7 N/m on each conductor. In this case, the wires are separated by a distance of 5.0 mm, which is equivalent to 0.005 m. Therefore, the force between the wires is 2x10-7 N/m * 0.005 m = 1x10-9 N per meter of separation.

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Answer:

[tex]P_{c[/tex]= 5.74W

[tex]P_{w}[/tex]=0.27 W

Explanation:

d= 2.6cm =>0.026m (for both the ceramic and the carpet )

Thermal conductivity of wool '[tex]k_{w}[/tex]'= [tex]k_{carpet}[/tex] = [tex]k_{wool}[/tex]=  0.04J/(sm °C)

Thermal conducticity of carpet '[tex]k_c}[/tex]' = 0.84 J/(sm °C)

Area 'A'= [tex]A_{carpet}[/tex]= [tex]A_{ceramic}[/tex]= 77.2cm²=> 77.2 x [tex]10^{-4}[/tex]m²

[tex]T_h}[/tex]=33.0°C

[tex]T_c}[/tex]=10.0°C

Average Power [tex]P_{avg}[/tex] is determined by dividing amount of energy'Q' by time taken for the transfer't':

[tex]P_{avg}[/tex] = Q/Δt

Due to conductivity, heat of flow rate will be P= dQ/dt

P=[tex]\frac{dQ}{dt}[/tex] = [tex]\frac{kA[T_{h}-T_{c} ]}{d}[/tex]

For CERAMIC:

[tex]P_{c[/tex]=[tex]\frac{k_{c} A[T_{h}-T_{c} ]}{d}[/tex] => [0.84 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026

[tex]P_{c[/tex]= 5.74W

For WOOL CARPET:

[tex]P_{w}[/tex]= [tex]\frac{k_{w} A[T_{h}-T_{c} ]}{d}[/tex]=> [0.04 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026

[tex]P_{w}[/tex]=0.27 W

Final answer:

The volume of the bubble just below the surface of the water is 16 cm3.

Explanation:

To solve this problem, we can use Boyle's law, which states that the pressure times the volume of a gas is constant if the temperature remains constant. In this case, the volume of the air bubble at the bottom of the water is 16 cm3 and the depth is 6.5 m. As the bubble rises, the pressure decreases, causing the volume to increase. We can set up a proportion to solve for the volume just below the surface of the water.

Using the formula P1V1 = P2V2, where P1 is the pressure at the bottom of the water, V1 is the initial volume, P2 is the pressure just below the surface, and V2 is the volume just below the surface, we can plug in the values and solve for V2.

Since the temperature of the air in the bubble doesn't change, we can assume that the pressure times the volume is constant. Therefore, we have:

P1V1 = P2V2

Using the given values, we have:

(1 atm)(16 cm3) = (?, just below the surface) V2

To find V2, we can rearrange the equation:

V2 = (P1V1) / P2

Since the pressure just below the surface is atmospheric (1 atm), we can substitute that value in:

V2 = (1 atm)(16 cm3) / (1 atm)

V2 = 16 cm3

Therefore, the volume of the bubble just below the surface of the water is 16 cm3.

To move a 32 kg crate with a coefficient of static friction of 0.57 with the floor, the force necessary to start the crate moving is 180 N after rounding to two significant figures.

To find the force necessary to start the crate moving, we need to use the concept of static friction. The force of static friction is given by the equation fs=μsN, where fs is the static friction, μs is the coefficient of static friction and N is the normal force. The normal force is equal to the product of the mass and gravitational acceleration (N = mg), where m is the mass and g is the acceleration due to gravity. Given that the mass of the crate is 32 kg and the coefficient of static friction is 0.57, we first calculate N = (32kg)(9.8m/s²) = 313.6 N.

Then, by substituting these values into the equation, we have fs = (0.57)(313.6 N) =178.85 N. Therefore, the force necessary to start the crate moving is 178.85 N, but we express the answer using two significant figures, so it is 180 N.

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Answer:

Explanation:

Given:

Number of turns [tex]N = 200[/tex]

Cross sectional area [tex]A = 10^{-3}[/tex] [tex]m^{2}[/tex]

Rate of increasing magnetic field [tex]\frac{dB}{dt} = 6[/tex] [tex]\frac{T}{s}[/tex]

From the faraday's law,

Induced emf is given by,

   [tex]\epsilon = -N\frac{d \phi}{dt }[/tex]

Where [tex]\phi =[/tex] magnetic flux

  [tex]\phi = AdB \cos(0)[/tex]          ( because angle between normal coil and field is zero)

Where [tex]A =[/tex] area of coil

Put the value of [tex]\phi[/tex] in above equation,

Here we neglect minus sign

   [tex]\epsilon = NA\frac{dB}{dt}[/tex]

   [tex]\epsilon = 200 \times 10^{-3} \times 6[/tex]

   [tex]\epsilon = 1.2[/tex] V

Therefore, 1.2 volt induced in coil

Final answer:

The motion of a simple pendulum is not strictly a simple harmonic motion. While it exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, the restoring force is not linearly proportional to the displacement.

Explanation:

The motion of a simple pendulum is not strictly a simple harmonic motion because it does not follow Hooke's law. Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke's law. However, a simple pendulum is governed by the law of conservation of energy and the restoring force is not directly proportional to the displacement.

While a simple pendulum exhibits some characteristics of simple harmonic motion, such as periodic motion and a restoring force, it deviates from true simple harmonic motion because the restoring force is not linearly proportional to the displacement. Instead, the restoring force for a simple pendulum is proportional to the sine of the displacement angle.

For example, when a pendulum is displaced to one side, it experiences a restoring force in the opposite direction. But the magnitude of the restoring force does not increase linearly with the displacement. Instead, it follows the relationship F = -mg sin(theta), where F is the restoring force, m is the mass of the pendulum bob, g is the acceleration due to gravity, and theta is the displacement angle.

Answer:

B.

Explanation:

I'm taking the test right now.

The final momentum of the system is 0.4kgm/s

According to the law conservation of momentum, the momentum of the system before the collision and after the collision remains consevred.

if  [tex]m_{1}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{1}[/tex] and [tex]v_{1f}[/tex] respectively, and

if  [tex]m_{2}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{2}[/tex] and [tex]v_{2f}[/tex] respectively

then,   [tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{1f} + m_{2}v_{2f}[/tex]

 Momentum after collision = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex][tex]=0.6*0.5+0.5*0.2[/tex]

   Momentum after collision  = 0.4 kgm/s

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The substance that can be separated into several different elements is air. The correct option is A.

Air, which is a mixture mainly composed of nitrogen and oxygen, among other gases, can thus be separated into its component elements. Substances like Iron (Fe), Hydrogen (H), and Nickel (Ni) are elements and cannot be broken down into simpler substances by ordinary chemical means.

Answer:

False.

Explanation:

Kinetic energy means it must move

Answer: I believe that it is A) True

Disclaimer: I'm not quite sure. We learned about this recently though so possibly I'm right. Good luck though!

Explanation:

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

[tex]\text{Mass of the child} = m = 25kg[/tex]

[tex]\text{Acceleration due to gravity} = g = 9.81m/s^2[/tex]

[tex]\text{Height lifted} = h = 0.80m (Upward)[/tex]

Work done to upward the object

[tex]W = mgh[/tex]

[tex]W = (25)(9.81)(0.8)[/tex]

[tex]W = 196.2J[/tex]

Horizontal Force applied while carrying 10m,

[tex]F = 0N[/tex]

[tex]W = 0J[/tex]

Height descended in setting the child down

[tex]h' = -0.8m (Downwoard)[/tex]

[tex]W = mgh'[/tex]

[tex]W = (25)(9.81)(-0.80)[/tex]

[tex]W = -196.2J[/tex]

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

To calculate the work done on the child during different parts of the trip and for the whole trip, we need to consider the weight of the child and the vertical distances involved. The work done in lifting the child up and setting the child back down is equal to the weight of the child multiplied by the vertical distance. The work done in carrying the child horizontally is zero.

In this problem, we need to calculate the work done on the child during different parts of the trip and for the whole trip. Work, W, is defined as the product of force and displacement, W = Fd. Since the child is being lifted vertically, the work done in lifting the child up is equal to the weight of the child multiplied by the vertical distance lifted, Wup = mgh. The work done in carrying the child horizontally is zero since the displacement is perpendicular to the force. Finally, the work done in setting the child back down is also equal to the weight of the child multiplied by the vertical distance lowered, Wdown = mgh. Therefore, the total work done on the child for the whole trip is the sum of the work done in lifting the child up and setting the child back down, Wtotal = Wup + Wdown.

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Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]

where

[tex]\mu_0[/tex] is the vacuum permeability

[tex]I_1[/tex] is the current in the 1st wire

[tex]I_2[/tex] is the current in the 2nd wire

r is the separation between the wires

In this problem

[tex]I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m[/tex]

Substituting, we find the force per unit length on the two wires:

[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N[/tex]

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

Answer:

[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]

Explanation:

We name M the larger mas, and m the smaller mass, so base on this we write the following conservation of momentum equation for the collision:

[tex]P_i=P_f\\M\,v_i -m\,v_i=(M+m)\,v_f\\(M-m)\,v_i = (M+m)\,v_f\\\\\frac{v_i}{v_f} = \frac{(M+m)}{(M-m)}[/tex]

We can write this in terms of what we are looking for (the quotient of masses [tex]\frac{m}{M}[/tex]:

[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}[/tex]

We use now the information about  Kinetic Energy of the system being reduced in half after the collision:

[tex]K_i=2\,K_f\\\frac{1}{2} (M+m)\,v_i^2= 2*\frac{1}{2} (M+m)v_f^2\\v_i^2=2\,v_f^2\\\frac{v_i^2}{v_f^2} =2[/tex]

We can combine this last equation with the previous one to obtain:

[tex]\frac{v_i}{v_f} = \frac{(1+\frac{m}{M} )}{(1-\frac{m}{M} )}\\(\frac{v_i}{v_f})^2 = \frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}\\2=\frac{(1+\frac{m}{M} )^2}{(1-\frac{m}{M} )^2}[/tex]

where solving for the quotient m/M gives:

[tex]\frac{m}{M} =(\frac{2}{\sqrt{2} -1+1}) ^{-1}\\\frac{m}{M} =0.171572875[/tex]

Answer: 1) use of light with longer wavelenght will increase the width of the central maxima

2) increasing the number of slit will increase the width of the central maxima

3) narrowing the slit will increase the central maxima.

Answer:

7.7 MJ

Explanation:

Let water specific heat be c = 0.004186 J/kgC and specific latent heat of vaporization be L = 2264705 J/kg

There would be 2 kinds of heat to achieve this:

- Heat to change water temperature from 78C to boiling point:

[tex]H_1 = mc\Delta t = 3.4 * 0.004186 * (100 - 78) = 0.313 J[/tex]

- Heat to turn liquid water into steam:

[tex]H_2 = mL = 3.4 * 2264705 \approx 7.7 \times 10^6J[/tex] or 7.7 MJ

So the total heat it would take is [tex]H_1 + H_2 = 7.7 \times 10^6 + 0.313 \approx 7.7 MJ[/tex]

Answer: seen in the explanations

Explanation:

Since no values for this problem is given, I will just give the formulas to solve the problem in mechanics of orthogonal plane cutting.

a) shear plane angle will be calculated using:

tan ϕ = r Cos α / 1- r sin α

Where :

r= chip thickness ratio

α = rake angle

ϕ = shear angle.

B) chip thickness

To get the chip thickness, we must calculate the chip thickness ratio using the formula:

Rt = Lc/ L

Where:

Lc = length of chip formed

L= uncut chip length.

Use the answer to solve for chip thickness with the formula;

Tc = t/ Rt

Where :

t = depth of cut

Rt = chip thickness ratio

C) shear strain for the operation will be solved using;

ε = cot β• + tan β•

Where:

ε = shear strain

β• = orthogonal shear angle.

Answer:

32.3 m/s

Explanation:

We can solve this problem by applying one of Newton's equations of motion:

[tex]v^2 = u^2 + 2gs[/tex]

where v = final velocity of falcon

u = initial velocity of falcon

g = acceleration due to gravity ([tex]9.8 m/s^2[/tex])

s = distance moved by falcon

From the question, we have that:

u = 13.5 m/s

s = initial height - new height = 49.0 - 5.0 = 44.0 m

Hence, to find the velocity when it has traveled 44 m towards the ground (5 m above the ground):

[tex]v^2 = 13.5^2 + (2 * 9.8 * 44)\\\\\\v^2 = 182.25 + 862.4\\\\\\v^2 = 1044.65\\\\\\v = \sqrt{1044.65} \\\\\\v = 32.3 m/s[/tex]

The velocity of the falcon at the instant when it is 5.0 m above the ground is 32.3 m/s

Answer:

The speed of the box at the top of the hill will be 5.693m/s.

Explanation:

The kinetic energy of the box at the bottom of the hill is

[tex]K.E = \dfrac{1}{2}mv^2[/tex]

putting in [tex]m =24kg[/tex] and  [tex]v = 12.1m/s[/tex] we get

[tex]K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J[/tex]

Now, the potential energy this box gains as it rises [tex]h =5.7m[/tex] up the hill is

[tex]P.E = mgh[/tex]

[tex]P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368[/tex]

Therefore, the energy left [tex]E_{left}[/tex] in the box at the top if the hill will be

[tex]E_{left} =K.E - P.E = 1756.92J-1368J\\[/tex]

[tex]\boxed{E_{left} = 388.92J}[/tex]

This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,

[tex]\dfrac{1}{2}mv_t^2= 388.92J[/tex]

putting in numbers and solving for [tex]v_t[/tex] we get:

[tex]\boxed{v_t = 5.693m/s.}[/tex]

Thus, the speed of the box at the top of the hill is 5.693m/s.

Answer:

v = -1.3 mph

Explanation:

       [tex]p_{init} = p_{final} (1)[/tex]

       [tex]p_{init} = m_{car} * v_{car} - m_{SUV} * v_{SUV}[/tex]

        [tex]p_{init} = 1t*34 mph - 3t*13 mph = - 5t*mph (2)[/tex]

        [tex]p_{final} =( m_{car} + m_{SUV} )* v_{final} (3)[/tex]

       [tex]v_{final} = \frac{p_{init}}{(m_{car} + m_{SUV} } = \frac{-5t*mph}{4t} = -1.3 mph[/tex]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

Answer:

Explanation:

 3000 Calorie = 3000 x 1000 calorie

1 calorie = 4.2 Joule

3000 x 1000 calorie = 4.2 x 3000 x 1000 J

= 12.6 x 10⁶ J

It is taken in one day

time period of one day = 24 x 60 x 60 second

= 86400 s

energy given by this intake in 86400s is 12.6 x 10⁶ J

power = energy / time

= 12.6 x 10⁶ / 86400 J/s

= 145.8 J/s or W

= 145 W.

Final answer:

The average power of energy intake for a person consuming 1,300 to 3,000 kcals per day translates to approximately 63 to 145 watts. This calculation is done by converting calories to joules and then dividing by the total number of seconds in a day.

Explanation:

In order to calculate the average power of the energy intake from food, we can convert the amount of energy consumed per day into watts. For a range of 1,300 to 3,000 kcal per day, we need to convert these values to joules, since 1 kcal = 4.184 kJ. Therefore, 1,300 kcal converts to 1,300 x 4.184 kJ = 5,439.2 kJ and 3,000 kcal to 3,000 x 4.184 kJ = 12,552 kJ.

Next, we divide each amount by the number of seconds in one day to find the average power in watts. There are 86,400 seconds in a day. Thus, the power for 1,300 kcal/day is 5,439.2 kJ / 86,400 s = approximately 63 watts, and for 3,000 kcal/day is 12,552 kJ / 86,400 s = approximately 145 watts.

The power range for 1,300 to 3,000 kcal/day is thus approximately 63 watts to 145 watts.

Answer:

a) less than Mercury's.

Explanation:

For orbital time period  of a planet  , the expression is

T² = 4π² R³ / GM

T is time period , R is radius of orbit , G is universal gravitational constant , M is mass of the star or sun

T² ∝ R³

As radius of orbit increases , time period increases . The given planet is making around a star whose mass is equal that  of  sun so M is same as sun .

The given planet has time period equal to .01 years which is less than that of Mercury and Venus , hence its R will be less than orbit of both of them or less than mercury's .

Answer:

v:

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Answer: hola

Explanation:спасибо за очки

Answer:

A) 128 rad/s

B) 3.87 m/s

C) 1317.17 m/s²

D) 20.79m

Explanation:

A) We are given the angular speed as 1220 rev/min. Now let's convert it to rad/s.

1220 (rev/min) x (2πrad/1rev) x (1min/60sec) = (1220 x 2π)/60 rad/s = 127.76 rad/s ≈ 128 rad/s

B) Formula for tangential speed is given as;

v = ωr

We know that ω = 128 rad/s

Also, r = 3.02cm = 0.0302m

Thus, v = 128 x 0.0302 = 3.87 m/s

C) Formula for radial acceleration is given as;

a_c = v²/r

From earlier, v = ωr

Thus, a_c = v²/r = (ωr)²/r = ω²r

On the rim, r = 8.04cm = 0.0804

a_c = 128² x 0.0804 = 1317.17 m/s²

D) We know that; distance/time = speed

Thus, distance = speed x time

D = vt

From earlier, v = ωr

Thus, D = ωrt

Plugging in the relevant values ;

D = 128 x 0.0804 x 2.02 = 20.79m

Any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.

For a simple pendulum, the period is given as  

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

This is also given as

[tex]\bold {T = 2\pi \sqrt{\dfrac m{k}}}[/tex]  

where  

T   = period of oscillation  

m   = mass of the pendulum  

L    = length  

g   = acceleration due to gravity  

k    = force constant

Equate these equations,

[tex]\bold {T = 2\pi \sqrt{\dfrac L{g}}} = \bold {T = 2\pi \sqrt{\dfrac m{k}}}\\\\\bold {\bold { \dfrac L{g} = \dfrac m{k}}} }\\\\\bold {\bold { \dfrac m{L} = \dfrac k{g}}} }[/tex]

So, any pendulum that will have the same period with mass, m, and length, L, must have the ratio of (L/g) and the ratio of its mass to force constant (m/k) must also be equal to this ratio.

To know more about the period of pendulum,

https://brainly.com/question/21630902

Answer:

Distance will be 2.81 m

Explanation:

Detailed explanation and calculation is shown in the image below.