Tracy (of mass 49 kg) and Tom (of mass 77 kg) are standing at rest in the center of the roller rink, facing each other, free to move. Tracy pushes off Tom with her hands and remains in contact with Tom’s hands, applying a constant force for 0.8 s. Tracy moves 0.4 m during this time. When she stops pushing off Tom, she moves at a constant speed. What is Tracy’s constant acceleration during her time of contact with Tom? Answer in units of m/s 2 .

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

Answer:

DC: 60V

AC: 15Vrms (21.2V amplitude)

Explanation:

Since we know the current values for each case, we can calculate voltage in both scenarios.

For DC voltage:

Vdc = Idc * R = 60mA * 1000 Ohms = 60Vdc

For AC voltage:

Vac = Iac * R = 15mA * 1000 Ohms = 15 Vac (rms) the amplitude of this signal would be Vmax = 15Vac * 1.414 = 21.2V

Answer:

The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]

(b) is correct option.

Explanation:

Given that,

Length = 12 cm

Mass = 200 g

Extend distance = 27 cm

Distance = 5 cm

Phase angle =0°

We need to calculate the spring constant

Using formula of restoring force

[tex]F=kx[/tex]

[tex]mg=kx[/tex]

[tex]k=\dfrac{mg}{x}[/tex]

[tex]k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}[/tex]

[tex]k=13.06\ N/m[/tex]

We need to calculate the time period

Using formula of time period

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

Put the value into the formula

[tex]T=2\pi\sqrt{\dfrac{0.2}{13.6}}[/tex]

[tex]T=0.777\ sec[/tex]

At t = 0, the maximum displacement was 5 cm

So, The equation of the time-dependent function of the position

[tex]x(t)=A\cos(\omega t)[/tex]

Put the value into the formula

[tex]x(t)=5\cos(2\pi\times f\times t)[/tex]

[tex]x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)[/tex]

[tex]x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)[/tex]

[tex]x(t)=5\cos(8.08t)[/tex]

Hence, The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]

To solve the problem it is necessary to apply energy conservation.

By definition we know that kinetic energy is equal to potential energy, therefore

PE = KE

[tex]mgh = \frac{1}{2}mv^2[/tex]

Where,

m = mass

g = gravitaty constat

v = velocity

h = height

Re-arrange to find h,

[tex]h=\frac{v^2}{2g}[/tex]

Replacing with our values

[tex]h=\frac{26^2}{2*9.8}[/tex]

[tex]h = 34.498\approx 34.5m[/tex]

Therefore the correct answer is C.

Answer:m=86.36 kg

Explanation:

Given

mass of tackler [tex]m_T=100 kg[/tex]

initial velocity of Tackler [tex]u_t=4.1 m/s[/tex]

Final velocity of combined system [tex]v=2.2 m/s[/tex]

Let m be the mass of receiver

conserving momentum

[tex]m\times 0+m_t\times u_t=(m+m_t)v[/tex]

[tex]0+100\times 4.1=(100+m)\cdot 2.2[/tex]

[tex]410=(100+m)\cdot 2.2[/tex]

[tex]100+m=186.36[/tex]

[tex]m=86.36 kg[/tex]

Answer:

Force in the rocket will be 4166.64 N

Explanation:

We have given mass of the rocket m = 3 kg

Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec

Initial speed u = 0 m/sec

Distance traveled s = 90 cm = 0.9 m

From third equation of motion we know that [tex]v^2=u^2+2as[/tex]

[tex]50^2=0^2+2\times a\times 0.9[/tex]

[tex]a=1388.88m/sec^2[/tex]

From newton's law we F = ma

So force will be [tex]F=1388.88\times 3=4166.64N[/tex]

So force in the rocket will be 4166.64 N

The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The average force that acted on the rocket is 166.8 Newtons.

The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The change in velocity is 50.0 m/s and the distance traveled is 90.0 cm, which is equivalent to 0.90 m.

So, the acceleration of the rocket is 50.0 m/s divided by 0.90 m, which gives us 55.6 m/s². Now, we can use Newton's second law to calculate the force: F = m * a, where F is the force, m is the mass, and a is the acceleration.

Plugging in the values, we get F = 3.00 kg * 55.6 m/s² = 166.8 N. Therefore, the average force that acted on the rocket is 166.8 Newtons.

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Answer:

(a) check attachment

(b)5 m/s²

Explanation:

Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv

(a) for answer check attachment.

(b) For the magnitude of the balls initial acceleration:

     Initial net force(f) = mg - upthrust

                                  = [tex]mg - (\frac{m}{p} )pg.g[/tex]

     acceleration (a) = [tex]Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²[/tex]

c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)

                                [tex]mg=6πnrv + upthrust[/tex]

d.) For the magnitude of terminal velocity:

                                             [tex]mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s[/tex]

e.) when the ball reaches terminal velocity, the acceleration is zero (0)

Answer:

5 m/s

Explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet is u.

mass of pendulum, M = 9.50 kg

height raised, h = 1.28 m

The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system

Let vf be the final velocity after the impact

KE = PE

1/2 (M + m) Vf^2 = (M + m) x g x h

vf^2 = 2 x 9.8 x 1.28 = 25.088

vf = 5 m/s

Thus, the final speed of the combined mass system after the impact s 5 m/s.  

Final answer:

The final speed of the combined masses after the collision can be found using the conservation of momentum and mechanical energy principles, resulting in a final speed of approximately 1.20 m/s.

Explanation:

The final speed Vf of the combined masses after the collision can be calculated using the principle of conservation of momentum and conservation of mechanical energy.

By equating the initial momentum of the system before the collision to the final momentum of the system after the collision, you can determine the final speed Vf.

In this case, the final speed Vf is approximately 1.20 m/s.

Answer:

4.5 x 10^6 miles

Calculations can be viewed on the snapshot attached to this reply.

Thanks

q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

To calculate the potential energy of the system of three charges, we use the formula:

PE = k * q1 * q2 / r12 + k * q2 * q3 / r23 + k * q3 * q1 / r31

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1, q2, and q3 are the charges of the three point charges (4.10 nC, -2.90 nC, and 2.00 nC, respectively)

r12, r23, and r31 are the distances between the three point charges (20.0 cm, 9.0 cm, and 11.0 cm, respectively)

Plugging in the values, we get:

PE = (8.98755 x 10^9 N·m^2/C^2) * (4.10 x 10^-9 C) * (-2.90 x 10^-9 C) / 0.20 m + (8.98755 x 10^9 N·m^2/C^2) * (-2.90 x 10^-9 C) * (2.00 x 10^-9 C) / 0.09 m + (8.98755 x 10^9 N·m^2/C^2) * (2.00 x 10^-9 C) * (4.10 x 10^-9 C) / 0.11 m

PE = 1.19 x 10^-6 J

Therefore, the potential energy of the system of three charges if q3 is placed at x = + 11.0 cm is 1.19 x 10^-6 J.

To make the potential energy of the system equal to zero, we need to place q3 at a point where the forces from q1 and q2 cancel each other out. This means that the vectors representing the forces must be equal and in opposite directions.

The force from q1 on q3 is given by:

F13 = k * q1 * q3 / r31^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1 is the charge of q1 (4.10 nC)

q3 is the charge of q3 (2.00 nC)

r31 is the distance between q1 and q3

The force from q2 on q3 is given by:

F23 = k * q2 * q3 / r23^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q2 is the charge of q2 (-2.90 nC)

q3 is the charge of q3 (2.00 nC)

r23 is the distance between q2 and q3

For the forces to cancel each other out, we need:

F13 = -F23

Substituting in the equations for the forces, we get:

k * q1 * q3 / r31^2 = -k * q2 * q3 / r23^2

Solving for r31, we get:

r31 = r23 * sqrt(q2/q1)

Substituting in the values, we get:

r31 = 0.09 m * sqrt(-2.90 nC / 4.10 nC)

r31 = 0.069 m

Therefore, q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

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Answer:

[tex]r=5.278\times 10^{-4}\ m[/tex]

Explanation:

Given that:

Now, form the mathematical expression of Kinetic Energy:

[tex]KE= \frac{1}{2} m.v^2[/tex]

[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]

[tex]v^2=4.2198\times 10^{11}[/tex]

[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]

from the relation of magnetic and centripetal forces we have the radius as:

[tex]r=\frac{m.v}{q.B}[/tex]

[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]

[tex]r=5.278\times 10^{-4}\ m[/tex]

The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

The magnetic force and centripetal force of the electron is calculated as follows;

qvB = mv²/r

qB = mv/r

r = mv/qB

where;

The kinetic energy of the electron is given as;

K.E = ¹/₂mv²

mv² = 2K.E

v² = 2K.E/m

v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)

v² = 4.215 x 10¹¹

v = √4.215 x 10¹¹

v = 6.5 x 10⁵ m/s

The radius of their orbit is calculated as follows;

r = mv/qB

r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)

r = 5.287 x 10⁻⁵ m

Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

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Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

[tex]n_{r}=1.572[/tex]

[tex]n_{b}=1.587[/tex]

We need to calculate the angle for red wavelength

Using Snell's law,

[tex]n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}[/tex]

Put the value into the formula

[tex]1.572\sin37=1\times\sin\theta_{r}[/tex]

[tex]\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})[/tex]

[tex]\theta_{r}=71.0^{\circ}[/tex]

We need to calculate the angle for blue wavelength

Using Snell's law,

[tex]n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}[/tex]

Put the value into the formula

[tex]1.587\sin37=1\times\sin\theta_{b}[/tex]

[tex]\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})[/tex]

[tex]\theta_{b}=72.7^{\circ}[/tex]

We need to calculate the angle between the red and blue light

Using formula of angle

[tex]\Delta \theta=\theta_{b}-\theta_{r}[/tex]

Put the value into the formula

[tex]\Delta \theta=72.7-71.0[/tex]

[tex]\Delta \theta=1.7^{\circ}[/tex]

Hence, The angle between the red and blue light is 1.7°.

The Hydrogen discharge lamp emits light of specific wavelengths which pass through a flint-glass prism, refract (or bend), and display a color spectrum due to dispersion. The red and blue lights have different refraction angles because of their different wavelengths. The distinct color bands indicate discrete energy transitions in the hydrogen gas.

Light emission and its properties are central to this question. The light emitted by a hydrogen discharge lamp, which contains hydrogen gas at low pressure, indicates that there are discrete energies emitted, thus producing light of specific wavelengths. These distinct wavelengths are a result of H2 molecules being broken apart into separate H atoms when an electric discharge passes through the tube.

When this light enters a prism, it refracts or bends, leading to a phenomenon called dispersion. This happens because not all colors or wavelengths of light are bent the same amount: the degree of bending depends on the wavelength of the light, as well as the properties of the material (in this case, flint-glass prism). For instance, red and blue light have different refraction angles due to their different wavelengths of 656 nm and 486 nm respectively.

So, in the context of a hydrogen discharge lamp, the two prominent wavelengths, 656 nm (red) and 486 nm (blue), will refract differently when passing through the flint-glass prism. Further, the separation (or dispersion) of light into these distinct colored bands as it exits the prism gives us a line spectrum - visual evidence of the discrete energy transitions happening at the atomic level in the hydrogen gas of the lamp.

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The 93 kg man will be moving at a speed of 0.161 m/s after catching the ball that he threw at a wall and rebounded back to him, based on the principle of conservation of momentum.

This question involves the principle of conservation of momentum. According to the law of conservation of momentum, the total momentum of the system (man and the ball) is conserved before and after the collision with the wall. The total initial momentum, before the ball is thrown, is zero as everything is at rest.

When the man throws the ball, the ball gains momentum in one direction, and to keep the total momentum zero, the man must gain an equal amount of momentum in the opposite direction. On returning, the ball has changed its direction, so it has lost the initial momentum and gained the same amount in the opposite direction.

The man catches the ball, meaning he gains the momentum that the ball brought back, causing him to move. To determine how fast he is moving, we set the final momentum of the system (which should still be zero) equal to the initial momentum.

To calculate the speed (v) of the man we'll use the equation: v = 2 × ball_mass × ball_speed / man_mass = 2 × 0.653 kg × 11.3 m/s / 93 kg = 0.161 m/s

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The man will be moving at approximately 0.078 m/s after he catches the ball. This is found using the conservation of momentum, considering the ball's momentum before and after it collides with the wall and bounces back to the man.

The student's question involves the conservation of momentum, a fundamental concept in physics. When the man throws the ball and catches it after it bounces off the wall, both the ball and the man are involved in an isolated system where external forces are negligible. As momentum is conserved, the velocity of the man after catching the ball can be calculated using the conservation of momentum principle.

To solve this problem, we first consider the momentum before and after the collision with the wall. The man throws the ball at 11.3 m/s. The direction of the ball's momentum changes after it bounces back, but the speed remains the same due to the 'elastic' nature of the collision as mentioned in the problem statement (no energy loss).

The initial momentum of the system (man and ball) is zero because they're both at rest initially. When the man throws the ball, the ball gains momentum in the forward direction, and the man gains momentum in the opposite direction. This momentum for the man (in the opposite direction) is what will be calculated.

The initial momentum of the ball and man is:

The final momentum after throwing the ball is:

By conservation of momentum:

Initial momentum = Final momentum

0 = -0.653 kg * 11.3 m/s + 93 kg * v

Solving for v gives:

v = (0.653 kg * 11.3 m/s) / 93 kg

v ≈ 0.078 m/s

This is the speed of the man after catching the ball, moving opposite to the direction in which he threw the ball.

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The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

[tex]v=f\lambda[/tex]

where

v is the speed

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for [tex]\lambda[/tex], we find the wavelength:

[tex]\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m[/tex]

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Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, [tex]V_{p} = 120\ V[/tex]          (rms voltage)

Voltage at secondary, [tex]V_{s} = 13000\ V[/tex]  (rms voltage)

Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]

where

[tex]N_{p}[/tex] = No. of turns in primary

[tex]N_{s}[/tex] = No. of turns in secondary

[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108

(b) The power supplied to the line is given by:

Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]

(c) The current rating that the fuse should have is given by:

[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]

[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]

[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]

So torque on the rocket will be 1.11475 N -m

The work function of the metal can be calculated using Einstein's photoelectric equation and the given information about the wavelength of light, stopping potential, Planck's constant, and speed of light. The initial step is to calculate the energy of the photon; thereafter, calculate the kinetic energy of the photoelectron with the stopping potential, and finally find the work function by subtracting the kinetic energy from the photon energy.

To calculate the work function of the metal, we need to first figure out the energy of the photon hitting the metal surface. This can be found using Einstein's photoelectric equation: E = hc/λ. Here, h is Planck's constant (6.626 x 10-34 J · s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the light (350 nm or 350 x 10-9 m).

After calculating the energy of the photon, we can determine the energy required to eject the photoelectrons, known as the stopping potential, which in this case is 0.500 V. Note this is the extra energy required to stop the photoelectron and is equal to the kinetic energy of the photoelectron. Hence, 0.500V = 0.500 eV because 1V = 1eV for a single electron with charge e. To convert this to joules, we use the conversion 1eV = 1.60 x 10-19J.

Finally, in order to find the work function, we use Einstein's photoelectric equation again which states that the energy of the incident photon is equal to the work function (φ) plus the kinetic energy of the photoelectron. That is, E = φ + KE. From this, we can rearrange the equation to solve for φ, the work function of the metal where: φ = E - KE.

By substituting the appropriate values into this equation, the work function of the metal can be found out. But remember, work function is a property of the metal only and is independent of the incident radiation. Also, it determines the threshold frequency or cutoff wavelength of the metal beyond which photoelectric effect will not occur.

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Answer:

2.27m/s

Explanation:

The coefficient of sliding friction = 0.38, the incline angle = 60o

Coefficient of sliding friction = frictional force/ force of normal (fn)

The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity

Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m

Force tending to push the body down the incline plane = mgsin60 = 8.496m

Net force on the body = force pushing downward - Frictional force

Net force = 8.496m - 1.8639m = 6.632m

Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled

1/2mv^2 = 6.632m * 0.387in meters

Cancel mass on both side leaves 1/2v^2 = 6.632*0.387

V^2 = 2.566584*2 = 5.133

V = √5.133 = 2.27m/s

The speed of the block after it has traveled a distance of 38.7 cm is 1.2 m/s.

The speed of the block is calculated by applying the following formula.

Using work energy theorem;

The work done by the force of friction is equal to the change in kinetic energy of the box.

W = K.E

μmgd cosθ  = ¹/₂mv²

μgdcosθ = ¹/₂v²

v² = (2gd cosθ)μ

v = √ (2gd cosθμ)

where;

v = √ (2gd cosθμ)

v = √ (2 x 9.8 x 0.387  x cos(60) x 0.38)

v = 1.2 m/s

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Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate [tex]m=100 kg[/tex]

Crate slows down in [tex]s=1.5 m[/tex]

initial speed [tex]u=1.77 m/s[/tex]

inclination [tex]\theta =30^{\circ}[/tex]

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

[tex]W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2[/tex]

[tex]W_{gravity}=mg(0-h)=mgs\sin \theta [/tex]

[tex]W_{gravity}=-mgs\sin \theta[/tex]

[tex]W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N[/tex]

change in kinetic energy[tex]=\frac{1}{2}\times 100\times 1.77^2=156.64 J[/tex]

[tex]W_{friction}=156.64+735=891.645[/tex]

(b)Coefficient of sliding friction

[tex]f_r\cdot s=W_{friciton}[/tex]

[tex]891.645=f_r\times 1.5[/tex]

[tex]f_r=594.43 N[/tex]

and [tex]f_r=\mu mg\cos \theta [/tex]

[tex]\mu 100\times 9.8\times \cos 30=594.43[/tex]

[tex]\mu =0.7[/tex]

Final answer:

To answer part (a), calculate the work done by friction using the equation work = force × distance. Calculate the change in kinetic energy of the crate using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2 and subtract it from the work done by friction. To answer part (b), use the equation frictional force = coefficient of friction × normal force and calculate the normal force using the equation weight = mass × acceleration due to gravity.

Explanation:

To answer part (a), we need to calculate the work done by friction. The work done by friction can be calculated using the equation work = force × distance. In this case, the force is the frictional force and the distance is the distance traveled by the crate. The work done by friction is equal to the change in kinetic energy of the crate, which can be calculated using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2. Subtracting the change in kinetic energy from the work done by friction will give us the energy dissipated by friction.

To answer part (b), we can use the equation frictional force = coefficient of friction × normal force. The normal force is equal to the weight of the crate, which can be calculated using the equation weight = mass × acceleration due to gravity. By substituting the known values into these equations, we can find the coefficient of sliding friction.

Answer:

The speed of the car was 28 m/s.

Explanation:

Hi there!

The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:

W = ΔKE = final kinetic energy - initial kinetic energy

In this case, the final kinetic energy is zero, then:

W = - initial kinetic energy

Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:

Wfr = initial kinetic energy

The work done by friction is calculated as follows:

Wfr = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction

Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).

Then the friction force can be written as follows:

Fr = m · g · μ

Where:

m = mass of the car.

g =  acceleration due to gravity (9.8 m/s²)

The work done by friction will be:

W = m · g · μ · d

The equation of kinetic energy is the following:

KE =  1/2 · m · v²

Where

m = mass of the car.

v = speed.

Then:

W = KE

m · g · μ · d = 1/2 · m · v²

2 · g · μ · d = v²

2 · 9.8 m/s² · 0.40 · 98 m = v²

v = 28 m/s

The speed of the car was 28 m/s.

Answer:

Explanation:

When a spring attached with a block is stretched and released , the block starts moving under SHM. The elastic energy stored in it initially at the time of initial stretch is  repeatedly converted into kinetic energy and vice - versa while the body keeps moving under SHM. So we can tell that the mechanical energy of the system remains constant.

In the whole process the velocity of the body keeps changing in terms of both magnitude and direction . It happens because a variable force acts on the body constantly towards the equilibrium point  so its momentum also keeps changing all the time

Answer:

therefore, the probability that an individual will have a cholesterol level greater than 60 ​mg/dL.= 0.27

Explanation:

given data

Normal distribution

mean cholesterol level μ= 51.6 ​mg/dL

Standard deviation σ=  14.3 ​mg/dL

x= 60 mg/dL

We have to find out P(x>60)

We Know that [tex]P(x>a) =P(Z>(a-\mu)/\sigma)[/tex]

therefore, [tex]P(x>60) =P(Z>(60-51.3)/14.3)[/tex]

= P(Z>0.61)

= 1 - P(Z<0.61)

= 1 - 0.7291

= 0.27

therefore, the probability that an individual will have a cholesterol level greater than 60 ​mg/dL.= 0.27

To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we calculate the z-score and find the corresponding probability using a Z-table or calculator.

To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we need to use the z-score formula. The z-score is calculated by subtracting the mean from the value of interest and dividing it by the standard deviation. In this case, the z-score is (60 - 51.6) / 14.3 = 0.589. Using a Z-table or a calculator, we can find that the probability of a z-score greater than 0.589 is approximately 0.279.

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Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m =  mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3  (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as

[tex]P = P_{atm} +\rho gh[/tex]

Where,

[tex]P_{atm} =[/tex] Atmospheric Pressure

[tex]\rho =[/tex]Density, water at this case

g = Gravity

h = Height

The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.

Our values are given as

[tex]P_{atm} = 1.013*10^5Pa[/tex]

[tex]\rho = 1000Kg/m^3[/tex]

[tex]g = 9.8m/s^2[/tex]

[tex]h = 0.78m[/tex]

Replacing at the equation we have,

[tex]P = P_{atm} +\rho gh[/tex]

[tex]P = 1.013*10^5 +(1000)(9.8)(0.78)[/tex]

[tex]P = 108944Pa[/tex]

[tex]P = 0.1089Mpa[/tex]

Therefore the pressure inside the hose is 0.1089Mpa

Answer:

37.11231 m/s

Explanation:

u = Initial velocity

v = Final velocity = 0

s = Displacement = 78 m

g = Acceleration due to gravity = 9.81 m/s²

[tex]\mu[/tex] = Coefficient of kinetic friction = 0.9

Acceleration is given by

[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s[/tex]

The initial speed of that car is 37.11231 m/s

Answer:

 x = 6.58 m

Explanation:

For this exercise let's use the concept of conservation of mechanical energy. Let's look for energy at two points the highest and when the spring is compressed

Initial

    Em₀ = U = m g and

Final

    [tex]E_{mf}[/tex] = ke = ½ k x2

As there is no friction the mechanical energy is conserved

    Em₀ = [tex]E_{mf}[/tex]

     m g y = ½ k x²

Let's use trigonometry to face height

     sin θ = y / L

     y = L sin θ

     x = √ 2mg (L synth) / k

     x = √ (2 11.8 9.8 2.17 sin35.6 / 3.11 104)

      x = 6.58 m

Answer:

The number of available energy is [tex]4.820\times10^{45}[/tex]

Explanation:

Given that,

Energy [tex]E=4.9\times10^{-4}\ J[/tex]

Temperature = 2.7 K

Energy states per unit volume [tex]dE=4\times10^{-5}\ eV[/tex]

We need to calculate the number of available energy

Using formula of energy

[tex]N=g(E)dE[/tex]

[tex]N=\dfrac{8\pi\times E^2 dE}{(hc)^3}[/tex]

Where, h = Planck constant

c = speed of light

E = energy

Put the value into the formula

[tex]N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]

[tex]N=4.820\times10^{45}[/tex]

Hence, The number of available energy is [tex]4.820\times10^{45}[/tex]

To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.

The pressure would be defined as

[tex]P = P_0 + P_w +P_m[/tex]

Where,

[tex]P_0 =[/tex] Standard atmosphere pressure

[tex]P_W =[/tex]Pressure of Water

[tex]P_m =[/tex]Pressure of mercury

Therefore we have that

[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]

Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.

Replacing with our values we have that,

[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]

[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]

Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa

The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.

A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:

[tex]\frac{0.25m}{2} = 0.13 m[/tex]

We can calculate the pressure at the bottom of the cylinder (P) using the following expression.

[tex]P = P_{atm} + P_{Hg} + P_w[/tex]

where,

We can calculate the pressure exerted by each liquid (P) as:

[tex]P = \rho \times g \times h[/tex]

where,

The pressure exerted by the column of Hg is:

[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]

The pressure exerted by the column of water is:

[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]

The pressure at the bottom of the cylinder is:

[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]

The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.

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The maximum mass of phosphoric acid that can be formed is 34.2 g.

The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex]

The excess reagent that remains after the reaction is complete is 3.76 g.

Phosphoric acid (H3PO4), also known as orthophosphoric acid, is the most important oxygen acid of phosphorus.

It is a crystalline solid and is used to generate fertilizer phosphate salts.

Given,

The balanced chemical equation is

[tex]P_4O_1_0 + 6 H_2O =4 H_3PO_4[/tex]

The molar mass of the following

[tex]\bold{P_4O_1_0}[/tex] = 284 g

[tex]\bold{H_2O}[/tex] = 16 g

[tex]\bold{H_3PO_4}[/tex] = 98 g

Calculating the number of moles

[tex]\dfrac{\bold{P_4O_1_0}}{284 g} = 0.0873\; mol/P_4O_1_0[/tex]

[tex]\dfrac{\bold{13.2\;g\;H_2O}}{18 g} = 0.733\; mol/P_4O_1_0[/tex]

From equation, 1 mol of [tex]\bold{P_4O_1_0}[/tex] reacts with 6 mol of water, then for 0.0873 mol of [tex]\bold{P_4O_1_0}[/tex] will react

[tex]0.0873\; mol\; P_4O_1_0 \times 6 mol H_2O/ 1 molP_4O_1_0 = 0.524\; mol\; H_2O[/tex]

The number of moles of water remaining are:

0.733 mol - 0.524 mol = 0.209 mol

Convert the amount of water into mass unit

0.209 mol · 18 g / mol = 3.76 g

The number of moles of water remaining is 3.76 g

Assuming that the reaction has a yield of 100%

Since 1 mol of [tex]\bold{P_4O_1_0}[/tex]  produces 4 mol of phosphoric acid, 0.0873 mol [tex]\bold{P_4O_1_0}[/tex] will produce:

[tex]0.0873\; mol\; P_4O_1_0 \times 4 mol H_3PO_4/ molP_4O_1_0 = 0.349\; mol\; of\;phosphoric\;acid[/tex]

Then, the maximum mass of phosphoric acid that can be formed is

[tex]0.349\; mol\; H_3PO_4 \times 98\;g/ mol = 34.2\; g[/tex]

Thus, The maximum mass of phosphoric acid is 34.2 g, The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex], and the excess reagent that remains after the reaction is complete is 3.76 g.

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Final answer:

The maximum mass of phosphoric acid that can be formed is 35.6 grams, with diphosphorus pentoxide being the limiting reagent and 8.77 grams of water remaining as excess.

Explanation:

The maximum mass of phosphoric acid that can be formed is 35.6 grams. The formula for the limiting reagent is diphosphorus pentoxide (P4O6). After the reaction is complete, 8.77 grams of water will remain as the excess reagent.

Usually, to find the maximum mass of the product formed (phosphoric acid), and the mass of the excess reagent that remains after the reaction, it involves determining the limiting reagent by calculating the moles of reactants and using stoichiometry. The limiting reagent is the reactant that will be completely used up first during the chemical reaction, limiting the amount of product formed.

Answer

given,

height of aquarium = 6 m

Depth of fresh water = D = 1.50 m

horizontal length of the aquarium(w) = 8.40 m

total force increased when liquid is filled to depth = 4.30 m

g = 9.81 m/s²

ρ = 998 Kg/m³

force in the aquarium.

dF = PdA

[tex]F = \int PdA[/tex]

[tex]F = \int \rho\ g\ y\ (wdy)[/tex]

[tex]F = \rho\ g\ w \int ydy[/tex]

[tex]F = \rho\ g\ w\dfrac{y^2}{2}[/tex]

[tex]F = \dfrac{\rho\ g\ w\ y^2}{2}[/tex]

At D = 1.5 m

[tex]F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}[/tex]

F = 9.08 x 10⁴ N

At D = 4.30 m

[tex]F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}[/tex]

F = 7.46 x 10⁵ N

Total force on the wall increased by

ΔF  = 74.6 x 10⁴ - 9.08 x 10⁴

ΔF  = 65.52 x 10⁴ N