9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

Calculate the work done by the student

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]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The net torque  [tex]\tau = 21.95N \cdot m[/tex]

Explanation:

   From the question we are told that

      The length of each side is  [tex]L = 2.5 m[/tex]

       The first force is  [tex]F_1 = 18N[/tex]

        The second force is  [tex]F_2 = 20 N[/tex]

         The third force is [tex]F_3 = 11N[/tex]

The free body diagram for the question is shown on the second uploaded image

  Generally torque is mathematically represented as

       [tex]\tau = r * F[/tex]

 Where [tex]\tau[/tex] is the torque

             r is  the length from the rotating point to the point the force is applied, this is also the radius of the circular path made

             F is the force causing the rotation.

looking at the free body diagram we can deduce that L is the diameter of the circular path made as a result of toque

       Now  for the torque due to force [tex]F_1[/tex]

               [tex]\tau_1 = - F_1 * r_1[/tex]

The negative sign is because the direction of  [tex]F_1[/tex]  is clockwise

     =>  [tex]\tau_1 = - F_1 * \frac{L}{2}[/tex]

Substituting value

           [tex]\tau_1 = - 18 * \frac{2.5}{2 }[/tex]

            [tex]\tau_1 = - 22.5 N \cdot m[/tex]

The torque as a result of the second force is  mathematically evaluated as

            [tex]\tau_2 = F_2 * r_2[/tex]

            [tex]\tau_2 = F_2 * \frac{L}{2}[/tex]

                [tex]= 20 * \frac{2.5}{2}[/tex]

                 [tex]\tau_ 2 = 25 \ N \cdot m[/tex]

The torque as a result of the third  force is  mathematically evaluated as

            [tex]\tau_3 =r_3 (F_3 sin \theta + F_3 cos \theta )[/tex]

            [tex]\tau_3 = \frac{L}{2} (F_3 sin \theta + F_3 cos \theta )[/tex]

Where the free body diagram [tex]\theta = 45^o[/tex]

                [tex]\tau_3 = \frac{2.5}{2} (11 * sin (45) +11 cos (45) )[/tex]

                 [tex]\tau_ 3 = 19.45 \ N \cdot m[/tex]

The net torque the mathematically

       [tex]\tau = \tau_1 + \tau_2 + \tau_3[/tex]

substituting value

       [tex]\tau = -22.5 + 25 + 19.45[/tex]  

           [tex]\tau = 21.95N \cdot m[/tex]

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:

1.41*10^14 N

Explanation:

Given that

Weight of the person on earth, W = 665

Radius of the neutron star, r = 8 km

Mass of the sun, M = 1.99*10^30 kg

Gravitational constant, G = 6.67*10^-11 Nm²/kg²

Acceleration due to gravity in earth, g = 9.8 m/s²

Weight on earth is given by

W = mg, so that, mass m would be

m = W/g

m = 665 / 9.8

m = 67.86 kg

The mass on earth is 67.86 kg

Weight on the neutron star is then

W = F = GmM/r²

F = (6.67*10^-11 * 67.86 * 1.99*10^30) / 8000²

F = 9*10^21 / 6.4*10^7

F = 1.41*10^14 N

Thus, the the weight in the neutron star is 1.41*10^14 N

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

= mass * gravity * distance

= 1000 * 9.81 * 10  

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J  

Work done by cable = PE +KE  

108,100 J = KE + 98,100 J  

KE = 10,000 J

= 10kJ

=

Answer:

A)Work done by gravity = -98 Kj

B) W_tension = 108 Kj

C) Final kinetic energy Kf = 10 Kj

Explanation:

We are given;

mass; m = 1000kg

Upward acceleration; a = 1 m/s²

Distance; d = 10m

I've attached a free body diagram to show what is happening with the elevator.

A) From the image i attached, the elevators weight acting down is mg.

While T is the tension of the cable pulling the elevator upwards.

Now, we know that,

Work done = Force x distance

Thus, W = F•d

We want to calculate the work done by gravity;

From the diagram I've drawn, gravity is acting downwards and in am opposite direction to the motion having an upward acceleration.

Thus, Force of gravity = - mg

So, Work done by gravity;

W_grav = - mgd

W_grav = -1000 x 9.8 x 10 = -98000 J = -98 Kj

B) Again, W = F.d

W_tension = T•d

Let's find T by summation of forces in the vertical y direction

Thus, Σfy = ma

So, T - mg = ma

Thus, T = ma + mg

T = m(a + g)

Plugging in values,

T = 1000(1 + 9.8)

T = 1000 x 10.8 = 10800 N

So, W_tension = T•d = 10800 N x 10m = 108000 J = 108 Kj

C) From work energy theorem,

Net work = change in kinetic energy

Thus, W_net = Kf - Ki

Where Kf is final kinetic energy and Ki is initial kinetic energy.

Now since the elevator started from rest, Ki = 0 because velocity at that point is zero.

Thus, W_net = Kf - 0

W_net = Kf

Now, W_net is the sum of work done due to gravity and work done due to another force.

Thus, in this case,

W_net = W_grav + W_tension

W_net = -98000 J + 108000 J

W_net = 10000J = 10Kj

So,since W_net = Kf

Thus, Final kinetic energy Kf = 1000J

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: their final velocity will be zero

Explanation:

Since they have equal masses m,

And their velocity is equal but in opposite direction u and -u, then,

mu + (-mu) = 2mv

0 = 2mv

Which implies that V = 0

Answer:

Explanation:

Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field

a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.

b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will  exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.

Situations when electron is affected by an external magnetic field but not by an external electric field

There is no such situation in which electric field will not affect an electron . It will always affect an electron .

Final answer:

In certain conditions, an electron is affected by an external electric field but not by an external magnetic field. Conversely, an electron can be influenced by an external magnetic field but not by an external electric field.

Explanation:

An electron will be affected by an external electric field but will not be affected by an external magnetic field when it is at rest in a magnetic field as it experiences no force.

It is possible for an electron to be affected by an external magnetic field but not by an external electric field, as seen in the case of a moving charged particle forming a magnetic field around wires carrying electrical currents.

Answer:

Explanation:

The mass difference is

Δm = m2 - m1

56.936296 u - 56.935399 u

= 0.000897‬ u

The energy released by the electron-capture decay

E = Δmc²

  ( 0.000897‬ u) c² ( 931. 5 MeV /c²÷ 1 u)

= 0.8355555 MeV

Answer:

Explanation:

Polarization In this case angle of incidence is not equal to angle of polarization, hence reflected light is partially polarized and transmitted light is also partially polarized.  by reflection is explained by Brewster's law,  

According to this when unpolarized light incident on glass plate at an angle is called as angle of polarizing the reflected light is plane polarized, and transmitted light is partially polarized. The plane of vibration of polarized light is having plane of vibrations perpendicular to plane of incidence.

When two sheets of polarizing material with perpendicular axes are placed together, no light passes through. However, if a third sheet of polarizing material is placed between the first two sheets at an angle, some light can pass through the combination.

When two sheets of polarizing material are placed with their polarizing axes at right angles to each other, no light passes through the combination because the second sheet blocks the light that is polarized by the first sheet. However, if a third sheet of polarizing material is placed between the first two sheets, at an angle between 0° and 90° to the axis of the first sheet, some light can pass through the three-sheet combination.

This is because the third sheet allows a fraction of the previously blocked light to pass through. As the angle between the first and third sheets approaches 90°, more light is transmitted by the combination.

For example, in the case of Figure 27.41, when the axes of the first and second filters are aligned (parallel), all of the polarized light passed by the first filter is also passed by the second filter. When the second filter is rotated to make the axes perpendicular, no light is passed by the second filter.

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

The amplitude of the wave is 6.0 cm, the wavelength is 100π cm, the frequency is approximately 0.637 Hz, the speed is approximately 200π cm/s, the direction of propagation is the positive x-direction, and the maximum transverse speed of a particle in the string is 120π cm/s.

The given equation representing a transverse wave along a very long string is y = 6.0 sin(0.020px - 4.0pt). We can extract various information from this equation:

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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.

16.7 x 10²V

The work-energy theorem states that the change in kinetic energy of a particle, Δ[tex]K_{E}[/tex], results in work done by the particle, W. i.e;

Δ[tex]K_{E}[/tex] = W                     ------------------(i)

But:

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]mv² -

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]m [v² - u²]    ------------------(ii)

Where;

m = mass of particle (proton in this case) = 1.67 x 10⁻²⁷kg

v = final velocity of the particle = 6 x 10⁵m/s

u = initial velocity of the particle = 2 x 10⁵m/s

Substitute these values into equation (ii) as follows;

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(6 x 10⁵)² - (2 x 10⁵)²]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 x 10¹⁰) - (4 x 10¹⁰)]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 - 4) x 10¹⁰]

Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [32 x 10¹⁰]

Δ[tex]K_{E}[/tex] = (1.67 x 10⁻²⁷) [16 x 10¹⁰]

Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷J

Also:

W = qΔV         ----------------(iii)

Where;

q = charge of the particle (proton) = 1.6 x 10⁻¹⁹C

ΔV = change in potential difference of the particle = V (from the question)

Substitute these values into equation (iii) as follows;

W = 1.6 x 10⁻¹⁹ x V        

W = 1.6 x 10⁻¹⁹V         -------------------(iv)

Now:

Substitute the values of Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷ and W in equation(iv) into equation (i)

26.72 x 10⁻¹⁷ = 1.6 x 10⁻¹⁹V

Solve for V;

V = (26.72 x 10⁻¹⁷) / (1.6 x 10⁻¹⁹)

V = 16.7 x 10²V

Therefore, the potential difference through which the proton fell was 16.7 x 10²V

The potential contrast between different locations represents the work or energy dissipated in the transmission of the unit amount of voltage from one point to another.

Following are the calculation of the potential difference:

Change in energy [tex], e^{v}=\frac{1}{2} \ m(v_2^2-v_1^2)[/tex]

[tex]1.602 \times ^{-19}=\frac{1}{2} \times 1.67 \times 10^{-27} (6^2-2^2)\times 10^{10}\\\\[/tex]

[tex]\to v \times 1.602 \times 10^{-19}=0.835 \times 10^{-17}\times 32\\\\[/tex]

[tex]\to v = \frac{0.835 \times 10^{-17}\times 32}{1.602 \times 10^{-19}}\\\\[/tex]

        [tex]= \frac{0.835 \times 10^{2}\times 32}{1.602 }\\\\ = \frac{26.72 \times 10^{2} }{1.602 }\\\\=16.67\times 10^{2}[/tex]

Therefore the final answer is "[tex]\bold{16.67 \times 10^2}[/tex]".

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Answer:

W=-1881.6J

Explanation:

we have that the change in the mass is

[tex]\frac{dm}{dt}=-c\\m(0)=60kg\\m(8)=60kg-6kg=54kg[/tex]

by solving the differential equation and applying the initial conditions we have

[tex]\int dm=-c\int dt\\m=-ct+d\\m(0)=-c(0) + d=60 \\m(8)=-8c+d=54[/tex]

by solving for c and d

d=60

c=0.75

The work needed is

W = m(t) gh

by integrating we have

[tex]dW_T= gh\int dm \\\\W_T=gh\int_0^8 -0.75dt\\\\W_T=(9.8\frac{m}{s^2})(32m)(-0.75(8))=-1881.6J[/tex]

hope this helps!!

Answer:

b. There is no net force on Lindsay.

Explanation:

Since Lindsay manages to stay stationary then this means that either there is no force acting on Lindsay or all the forces acting on Lindsay are balanced and produces no net force on Lindsay with respect to the bus.

Answer:

No net force acting on Lindsay.              

Explanation:

Solution:-

- Lets assume the mass of the bus to be = M

- Assume mass of Lindsay = m

- The initial speed of bus and lindsay was vi = 0 m/s

- We will consider the system ( Lindsay + Bus ) to be isolated with no fictitious unbalanced forces acting on the system.

- Since, the system can be isolated the application of principle of conservation of linear momentum is completely valid.

- We will assume the direction of bus moving to be positive.

- The principle of conservation of momentum states:

                           Pi = Pf

                           vi*(m+M) = m*vL + M*vb

Where,

             vL : The velocity of Lindsay after bus starts moving

             vb : The velocity with which bus moves

Since,    vi = 0 m/s.

                           0 = m*vL + M*vb

                          vL = - ( M / m )*vb

- The lindsay moves in opposite direction of motion of bus. We will apply the Newton's second law of motion on Lindsay:

                           Impulse = F*t = m*( vL - vi )

                           Impulse = F*t = m*( - ( M / m )*vb - 0 )

                           Impulse = F*t = -M*vb

                          F = -M*vb/t

- We see that a force ( F ) acts on Lindsay as soon as the bus starts moving. The negative sign shows the direction of force which is opposite to the motion of bus. So the force acts on Lindsay towards the back of the bus.      

- However, if we consider the system of ( bus + Lindsay ), the net force exerted on Lindsay remains zero as the impulse force ( F ) becomes an internal force of the system and is combated by the friction force ( Ff ) between Lindsay's shoes and the aisle floor.

- The magnitude of Friction force ( Ff ) is equivalent to the impulse created by force ( F ) by Newton's third Law of motion. Every action has its equal but opposite reaction. Hence,

                          Fnet = F + Ff

                          Fnet = F - F

                          Fnet = 0

- However, The force ( F ) created by the rate of change of momentum of bus must be less than Ff which is equivalent to weight of Lindsay:

                          Ff = m*g*us

Where,

             us: The coefficient of static friction.

- So for forces to remain balanced, with no resultant force then:

                         Ff ≥ F

                         m*g*us ≥ M*ab

                         us ≥ (M/m)*(ab/g)

- The coefficient of static friction should be enough to combat the acceleration of bus ( ab ).

Explanation:

answer and explanation is in the picture

Answer:

Both cylinders will have the same moment of inertia.

Explanation:

The moment of inertia of a rigid body depends on the distribution of mass around the axis of rotation, i.e at what radius from the axis how much mass is located. What the moment of of inertia DOES NOT depend in is the distribution of mass parallel to the axis of rotation. This means that two hollow cylinders of the same mass but with different lengths will have the same moment of inertia!

For completeness, the momentum of inertia of a hollow cylinder is

[tex]I \approx mR^2[/tex]

from which we clearly see that  [tex]I[/tex] only depends on the mass and the radius of the hollow cylinder and not on its height; Hence, both the wooden and the brass cylinders will have the same moment of inertia.

The two (2) hollow cylinders will have the same moment of inertia about their cylindrical center axis.

Moment of inertia is also referred to as the mass moment of inertia and it can be defined as a measure of the rotational inertia of an object or its resistance to angular acceleration about a reference axis.

Mathematically, the moment of inertia of a hollow cylinder is given by the formula:

[tex]I=mr^2[/tex]

Where:

From the above formula, we can deduce that the moment of inertia of a hollow cylinder is highly dependent on the distribution of mass around its axis of rotation (radius) rather than its height.

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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 final temperature is 21.531°c

Explanation:

Heat gained by the water = heat lost by the metal

Heat gained = (m)(c)(∆tita)

Where m = mass

C = specific heat capacity

∆tita = temperature change

X = equilibrium temperature

So ....

Heat gained by water

= 175*4.185*(x-20)

= 732.2x - 14644

Heat lost by metal

= 44.5*0.321*(100-x)

=1428.45 -14.2845x

So....

1428.45 - 14.2845x = 732.2x - 14644

1428.45+14644= 732.2x + 14.2845x

16072.45= 746.4845x

16072.45/746.4845= x

21.531 = x

The final temperature is 21.531°c

Answer:

Only 9% weaker

Explanation:

Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.

Answer:

The gravity at the altitude of 450 km above the earth's surface is (0.87 × the normal acceleration due to gravity); that is, 0.87g.

This shows that the gravity at this altitude isn't as weak as initially thought; it is still 87% as strong as the gravity on the surface of the earth.

The gravity is only 13% weaker at an altitude of 450 km above the earth's surface.

Explanation:

The force due to gravity that is usually talked about is basically the force of attraction between the planetary bodies such as the earth and objects on its surface.

The force is given through the Newton's gravitational law which explains that the force of attraction between two bodies is directly proportional to the product of the masses of both bodies and inversely proportional to the square of their distance apart .

Let the force of attraction between a body of mass, m on the surface of the earth and the earth itself, with mass M

F ∝ (Mm/r²)

F = (GMm/r²)

G = Gravitational constant (the constant of proportionality)

r = distance between the body and the earth, and this is equal.to the radius of the earth.

This force is what is now translated to force of gravity or weight of a body.

F = mg

where g = acceleration due to gravity = 9.8 m/s²

F = (GMm/r²) = (mg)

g = (GM/r²) = 9.8 m/s²

So, for a body at a distance that is 450 km above the earth's atmosphere, the distance between that body and the centre of the earth is (r + 450,000) m

Let that be equal to R.

R = (r + 450,000) m

Note that earth's radius is approximately 6400 km

r = 6400 km = 6,400,000 m

R = 6400 + 450 = 6850 km = 6,850,000 m

Normal acceleration due to gravity = 9.8 m/s²

9.8 = (GM/6,400,000²)

GM = 9.8 × 6,400,000²

Acceleration due to gravity at a point 450 km above the earth's surface

a = (GM/R²)

a = (GM/6,850,000²)

Note that GM = 9.8 × 6,400,000²

a = (9.8 × 6,400,000²) ÷ (6,850,000²)

a = 9.8 × 0.873 = 8.56 m/s²

a = 87% of g.

The gravity at this altitude above the earth's surface is (0.87 × the normal acceleration due to gravity)

Hope this Helps!!!

Answer:

When you hit a baseball with a bat, you are applying a force only in the time while the bat is in contact with the ball, after that, the horizontal speed of the ball is constant. (if you ignore the friction of the air)

This happens because of the second Newton's law:

Force is equal to mass times the acceleration or:

F = m*a

If we do not have a force, then we do not have acceleration.

This means that the change in position for a fixed amount of time is constant, so horizontal speed is constant.

Answer:

a) Фm = μ₀*I*n*π*N*R₁²

b) Фm = μ₀*I*n*π*N*R₂²

Explanation:

Given

n = number of turns per unit length of the solenoid

R₁ = radius of the solenoid

I = current passing through the solenoid

R₂ = radius of the circular coil

N = number of total turns on the coil

a) Фm = ?   If  R₂ > R₁

b) Фm = ?   If  R₂ < R₁.

We can use the formula

Фm = N*B*S*Cos θ

where

N is the number of turns

B is the magnetic field

S is the area perpendicular to magnetic field B.

The magnetic field outside the solenoid can be approximated to zero. Therefore, the flux through the coil is the flux in the core of the solenoid.

We know that the magnetic field inside the solenoid is uniform. Thus, the flux through the circular coil is given by the same expression with R2 replacing R1 (from the area).

a)  Then, the flux through the large  circular loop outside the solenoid (R₂ > R₁) is obtained as follows:

Фm = N*B*S*Cos θ

where

B = μ₀*I*n

S = π*R₁²

θ = 0°

⇒ Фm = (N)*(μ₀*I*n)*(π*R₁²)*Cos 0°

⇒ Фm = μ₀*I*n*π*N*R₁²

b) The flux through the coil when  R₂ < R₁ is

Фm = (N)*(μ₀*I*n)*(π*R₂²)*Cos 0°

⇒ Фm = μ₀*I*n*π*N*R₂²

Answer:

a) ∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]

b) ∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]

Explanation:

The magnetic field inside the solenoid is uniform and the flux in the coil is the same as the flux in the solenoid.

The magnetic field outside can be said to be zero which also means that the flux through the coil is the same as flux in the solenoid.

Thus, we can say

R2 = R1

Where:

N = total turns on coil

n = number of turns per unit length of solenoid

I = current

R1 = radius of solenoid

R2 = circular coil radius

a) The magnetic flux through the coil if R2 > R1 is giben as:

∅ [tex]= N(u_0 *n*I)(pi*(R_1 )^2) Cos 0[/tex]

Solving further, we have:

∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]

Here,

[tex] u_0*n*I [/tex] is the magnetic field

(Pi*R1²) is the area pependicular to magnetic field

N is the number of turns

b) The magnetic flux through the coil if R2 < R1

∅ [tex] = N(u_0*n*I)(pi*(R_2)^2)Cos 0 [/tex]

∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]

Here,

[tex] u_0*n*I [/tex] is the magnetic field

(Pi*R2²) is the area pependicular to magnetic field

N is the number of turns

Answer:

DeskLamp

Explanation:

You're average toaster uses so much more electricity than the average desk lamp. 2 slice toasters I believe uses about 1000 watts? And 4 slice uses 1600. The brightest lightbulbs tend to run on 100 watts so just imagine a lightbulb with 1600 watts. (There's a video of a man cranking the watts to a 1000 watt lightbulb up to millions until it pops)  Lightbulbs have a very far range in the amounts of energy they can take in. Here is a simplified answer-

On average- the toaster would use the most watts in terms of use

In other circumstances- The lightbulb would be the case.

So yeah,

It's Desklamps. They're really powerful.

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:

- 21.3 kJ

Explanation:

For the favorable reaction, ΔG₁ = + 9.2 kJ (ΔG > 0 for unfavorable reaction)

For the favorable reaction, ΔG₂ = - 30.5 kJ (ΔG < 0 for favorable reaction)

For the overall reaction, ΔG = ΔG₁ + ΔG₂ = + 9.2 kJ + (- 30.5 kJ)

ΔG = + 9.2 kJ - 30.5 kJ = - 21.3 kJ

Explanation:

Mass, m = 0.464 kg

Compression in the spring, x = 0.646 m

(a) The net force acting on the spring is given by :

[tex]kx=mg[/tex]

k is spring constant

[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{0.464\times 9.8}{0.646 }\\\\k=7.03\ N/m[/tex]

(b) The angular frequency of the spring mass system is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{7.03}{0.464 }} \\\\\omega=3.89\ rad/s[/tex]

The period of oscillation is :

[tex]T=\dfrac{2\pi}{\omega}\\\\T=\dfrac{2\pi}{3.89}\\\\T=1.61\ m/s[/tex]

(c) As the spring oscillates, its will reach to a height of 2(0.338) = 0.676 m

The spring constant is calculated to be 7.03 N/m using Hooke's Law. The period of oscillation for the block-spring system is found to be 1.74s utilizing the formula for simple harmonic motion. The height reached above the point of release by the block is 0.338m, based on the conservation of energy principle.

To find the spring constant, which can be denoted by k, we can use Hooke's Law (F = -kx). The force, F, in this case is the weight of the block (mass x gravity), mg = 0.464 kg x 9.8 m/s² = 4.5472 N. This force causes the spring to stretch 0.646 m. Therefore, k is calculated as F/x = 4.5472 N / 0.646 m = 7.03 N/m.

The period of oscillation, T, for a block-spring system executing simple harmonic motion is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Substituting the given values, T = 2π√(0.464 kg / 7.03 N/m) = 1.74s.

For the height above the point of release that the block will reach, again we come back to conservation of energy. At the point of release, the block possesses only elastic potential energy, 1/2kA², where A is the amplitude. This energy will be entirely converted into gravitational potential energy, mgh, at the block's highest point. Therefore, h = (1/2kA²)/mg. So, h = (1/2 * 7.03 N/m * (0.338 m) ²) / (0.464 kg * 9.8 m/s²) = 0.338 m, above the point of release.

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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:

Work in physics is done when a force causes displacement in the direction of the force. Lifting a pencil off a desk and compressing a spring are both examples of work being done because in both cases there is a force and a displacement.

In physics, work is defined as the product of force, displacement, and the cosine of the angle between them. It's represented by the formula W = Fd cos(θ). Let's apply this definition to two scenarios:

Considering the examples from the given reference, we understand that work requires both a force and a displacement. No work is done if either the force or the displacement is zero, or if the force is perpendicular to the direction of the displacement.

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Answer:

a) 3.6 J

b) 0.9 J

c) 0 J

d) 0 J

e) 2.7 J

Explanation:

Work = force F x diaplament in direction of force d

Given that,

Displacement in direction of push force d = 1.5 N

Push force = 2.4 N

Friction = 0.6 N

a) work done by push force = 2.4 x 1.5 = 3.6 J

b) word done by friction = -0.6 x 15 = -0.9 J (minus sign shows the work is opposite that done by the push force)

c) work done by normal force from table top ( force acting upwards to oppose gravity) = 0 J since there is no vertical displacement

d) work done by gravity (doward force) = 0 J since there is no vertical displacement

e) net work done on book = 3.6 J + (-0.9J) + 0 J + 0 J = 2.7 J

The work done by each force is calculated using the formula Work = Force × Distance. The net work done on the book is the sum of the work done by each force. (a) 3.6 J (b) 0.9 J (c) 0 J (d) 0 J (e) 4.5 J

To calculate the work done by each force, we use the formula:

Work = Force × Distance

(a) The work done by your push is:

Work = 2.40 N × 1.50 m = 3.6 J

(b) The work done by the friction force is:

Work = 0.600 N × 1.50 m = 0.9 J

(c) The normal force from the tabletop does no work because it is perpendicular to the direction of motion.

(d) The work done by gravity is zero since the book doesn't move vertically and gravity acts vertically.

(e) The net work done on the book is the sum of the work done by all forces:

Net Work = Work by your push + Work by friction + Work by normal force + Work by gravity = 3.6 J + 0.9 J + 0 J + 0 J = 4.5 J

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