A 1400 kg car moving at 6.6 m/s is initially traveling north in the positive y direction.
After completing a 90° right-hand turn to the positive x direction in 4.9 s, the inattentive operator drives into a tree, which stops the car in 430 ms.

What is the magnitude of the impulse on the car :

(a) due to the turn and (b) due to the collision?

What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision?

(e) What is the angle between the average force in (c) and the positive x direction?

The answer to this problem can be given through energy conservation as well as Newton's first law.

Newton's first law states that as long as there is no force exerted on a body, its movement will be constant or at rest. In this way the amount of linear movement of the satellite before the explosion will be equal to the sum of the movement generated in the pieces after the explosion. In the absence of external forces but the preservation of the internal force as the only one to act, these will not change the total momentum of the system.

To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.

By definition we know that the magnetic field is,

[tex]B = \mu_0 n I[/tex]

[tex]\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI[/tex]

[tex]\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}[/tex]

At the same tome we know that the induced voltage is defined as

[tex]\epsilon = \frac{\Phi}{dt}[/tex]

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

Replacing

[tex]\epsilon = A \mu_0 n\frac{dI}{dt}[/tex]

[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}[/tex]

PART A) Substituting with our values we have that

[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m\\[/tex]

Therefore there is not induced electric field at the center of solenoid.

PART B) Replacing the radius for 0.5cm

[tex]\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m[/tex]

Therefore the magnitude of the induced electric field at a point 0.5cm is [tex]7.9168*10^{-5}V/m[/tex]

The magnitude of the induced emf near the center of the solenoid is 0 V/m.

The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is 8 x 10⁻⁵ V/m.

The given parameters;

The magnitude of the induced emf near the center of the solenoid is calculated as follows;

[tex]B = \mu_0 nI\\\\\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \\\\\frac{dB}{dt} = (4\pi \times 10^{-7} \times 700 \times 36) = 0.032 \ T/s[/tex]

[tex]E = \frac{r}{2} (\frac{dB}{dt} )\\\\\ E = \frac{0}{2} (0.032)\\\\ E = 0 \ V/m[/tex]

The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is calculated as follows;

[tex]E = \frac{r}{2} (\frac{dB}{dr} )\\\\ E= \frac{0.5\times 10^{-2} }{2}( 0.032)\\\\E = 8\times 10^{-5} \ V/m[/tex]

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Answer:

ΔX = 0.0483 m

Explanation:

Let's analyze the problem, the car oscillates in the direction y and advances with constant speed in the direction x

The car can be described with a spring mass system that is represented by the expression

     y = A cos (wt + φ)

The speed can be found by derivatives

     [tex]v_{y}[/tex] = dy / dt

    [tex]v_{y}[/tex]  = - A w sin (wt + φ

So that the amplitude is maximum without (wt + fi) = + -1

      [tex]v_{y}[/tex]  = A w

X axis

Let's reduce to the SI system

     vₓ = 15 km / h (1000 m / 1 km) (1h / 3600s) = 4.17 m / s

As the car speed is constant

     vₓ = d / t

      t = d / v ₓ

      t = 4 / 4.17

      t = 0.96 s

This is the time between running two maximums, which is equivalent to a full period

     w = 2π f = 2π / T

     w = 2π / 0.96

     w = 6.545 rad / s

We have the angular velocity we can find the spring constant

     w² = k / m

    m = 1200 + 4 80

    m = 1520 m

     k = w² m

     k = 6.545² 1520

     k = 65112 N / m

Let's use Newton's second law

    F - W = 0

    F = W

    k x = W

    x = mg / k

Case 1  when loaded with people

   x₁ = 1520 9.8 / 65112

   x₁ = 0.22878 m

Case 2 when empty

   x₂ = 1200 9.8 / 65112

   x₂ = 0.18061 m

The height variation is

    ΔX = x₁ -x₂  

    ΔX = 0.22878 - 0.18061

    ΔX = 0.0483 m

Final answer:

Calculating the rise in the car's suspension requires determining the spring constant from the initial displacement due to an 80 kg person's weight and then using it to find the new displacement when that weight is removed.

Explanation:

The problem relates to simple harmonic motion and the effects of weight on a car's suspension system. Upon the removal of the combined mass of the four 80 kg people (total of 320 kg) from the 1200 kg car, we need to find out how the car body rises due to this change in load. First, let's calculate the spring constant (k) using Hooks law (F = kx), where F is the force and x is the displacement.

When an 80 kg person gets in, the displacement (x) is 1.20 cm or 0.012 m, and the force (F) due to their weight is their mass times gravity (F = mg = 80 kg \\times 9.8 m/s²). Next, to find the new displacement caused by the removal of 320 kg, we'll set up the equation kx = mg for both situations and solve for the new displacement (x).

Answer:

[tex]K=512J[/tex]

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass [tex]m_1[/tex] has an initial velocity [tex]v_{1i}[/tex] and a final velocity [tex]v_{1f}[/tex] and the block of mass [tex]m_2[/tex] has an initial velocity [tex]v_{2i}[/tex] and a final velocity [tex]v_{2f}[/tex] then the initial and final momentum of the system will be:

[tex]p_i=m_1v_{1i}+m_2v_{2i}[/tex]

[tex]p_f=m_1v_{1f}+m_2v_{2f}[/tex]

Since momentum is conserved, [tex]p_i=p_f[/tex], which means:

[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]

We know that the block is brought to rest by the collision, which means [tex]v_{2f}=0m/s[/tex] and leaves us with:

[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}[/tex]

which is the same as:

[tex]v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}[/tex]

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

[tex]v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s[/tex]

So kinetic energy of the bullet as it emerges from the block will be:

[tex]K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J[/tex]

Answer:

(a) [tex]\eta_{max}=37.895\%[/tex]

(b) [tex]\eta=79.75\%[/tex]

(c) Impossibly

Explanation:

Given:

(a)

Now the maximum theoretical efficiency of the engine:

[tex]\eta_{max}=\frac{T_H-T_L}{T_H}\times 100\% [/tex]

[tex]\eta_{max}=\frac{950-590}{950}\times 100\% [/tex]

[tex]\eta_{max}=37.895\%[/tex]

(b)

Actual efficiency of the heat engine:

[tex]\eta=\frac{Q_H-Q_L}{Q_H}\times 100\% [/tex]

[tex]\eta=\frac{790-160}{790}\times 100\% [/tex]

[tex]\eta=79.75\%[/tex]

(c)

This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.

The answers are:

(a) the maximum theoretical efficiency is 37.9%

(b) the actual efficiency is 79.7%

(c) The cycle is operating impossibly

The following data is provided to us:

The temperature of the source, [tex]T_1=950K[/tex]

The temperature of the sink, [tex]T_2=590K[/tex]

Heat absorbed by the engine, [tex]Q_1=790kJ[/tex]

Heat rejected by the engine, [tex]Q_2=160kJ[/tex]

(a) Now, the maximum efficiency of the engine is:

[tex]\eta _{max}=1-\frac{T_2}{T1}\\ \\\eta_{max}=1-\frac{590}{950}\\ \\\eta_{max}=0.379[/tex]

maximum efficiency ( [tex]\eta_{max}[/tex] ) = 37.9%

(b) actual efficiency:

[tex]\eta=1-\frac{Q_2}{Q_1}\\ \\\eta=1-\frac{160}{790}\\ \\\eta=0.797[/tex]

actual efficiency ([tex]\eta[/tex]) = 79.7%

(c) it is not possible for a Carnot engine to have more efficiency than the maximum possible efficiency.

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Answer: e. 37

Explanation:

When two vectors are added, the maximum and minimum posible values, happen when both vectors are aligned each other.

If both vectors aim in the same direction, the maximum value is just the arithmetic sum of their lengths, in this case, 60.

If they aim in opposite direction, the resultant is the substraction of their magnitudes, which yields 20.

Any other posible value (depending on the angle between vectors, which can span from 0º to 180º) must be between those values.

So, the only choice that fits within this interval, is 37.  

Answer:

0.0085 Wb

Explanation:

a = Side of square

r = Radius of circle

[tex]\phi_s[/tex] = Magnetic flux through square loop = [tex]6.68\times 10^{-3}\ Wb[/tex]

Magnetic flux is given by

[tex]\phi=BA[/tex]

For square

[tex]\phi_{s}=Ba^2[/tex]

The length of the square will be equal to the circumference of the circle

[tex]4a=2\pi r\\\Rightarrow r=\frac{2a}{\pi}[/tex]

For circle

[tex]\phi_{c}=B\pi r^2\\\Rightarrow \phi_{c}=B\pi \left(\frac{2a}{\pi}\right)^2\\\Rightarrow \phi_c=Ba^2\frac{4}{\pi}\\\Rightarrow \phi_c=\phi_s\frac{4}{\pi}\\\Rightarrow \phi_c=6.68\times 10^{-3}\frac{4}{\pi}\\\Rightarrow \phi_c=0.0085\ Wb[/tex]

The flux that passes through the circular loop is 0.0085 Wb

The magnetic flux that passes through the circular loop is 0.00848 Wb.

The magnetic flux is defined as the measurement of the total magnetic field which passes through a given area.

Given that the magnetic flux that passes through the square loop is 6.68 × 10-3 Wb.

Let's consider that r is the radius of the circle and a is the side of the square.

The magnetic flux through the square loop is given below.

[tex]\phi_s = Ba^2[/tex]

Where B is the magnetic field.

The length of the square will be equal to the circumference of the circle

[tex]4a = 2\pi r[/tex]

[tex]r = \dfrac {2a}{\pi}[/tex]

The magnetic flux through the circular loop is given below.

[tex]\phi = BA[/tex]

Where A is is the cross-sectional area of the circular loop.

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

[tex]\phi = B \pi (\dfrac {2a}{\pi})^2[/tex]

[tex]\phi = B\pi \times \dfrac {4a^2}{\pi^2}[/tex]

[tex]\phi = Ba^2 \dfrac {4}{\pi}[/tex]

[tex]\phi = \phi_s \dfrac {4}{3.14}[/tex]

[tex]\phi = 6.68 \times 10^{-3} \times 1.27[/tex]

[tex]\phi = 0.00848 \;\rm Wb[/tex]

Hence we can conclude that the magnetic flux that passes through the circular loop is 0.00848 Wb.

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To determine the maximum height h that the container can be lifted before bursting, we can use the formula ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container, g is the acceleration due to gravity, and h is the height difference.

To determine the maximum height h that the container can be lifted before bursting, we need to consider the difference in pressure inside and outside the container.

The maximum pressure difference that the container can withstand before bursting is given as ΔPmax = 2.26 × 105 Pa.

To calculate the maximum height h, we can use the formula: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid inside the container (assuming it is the same as seawater density, rhos = 1025 kg/m3), g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height difference.

Plugging in the values, we get ΔPmax = 1025 * 9.81 * h. Solving for h, we find h ≈ 231 meters.

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Answer:

b

Explanation:

The space shuttle, in circular orbit around the Earth, collides with a small asteroid which ends up in the shuttle's storage bay.

This form of collision is called inelastic collision. And inelastic collision momentum is conserved but the kinetic energy is not conserved. Hence the correct option is b. only momentum is conserved.

Answer:

b

Explanation:

This form of collision is called inelastic collision.

Answer:

Wavelength of light will be 632.73 nm

Explanation:

We have given width of the single slit [tex]d=1850nm-1850\times 10^{-9}m[/tex]

Angle is given as [tex]\Theta =20^{\circ}[/tex]

We have to find the wavelength of light for first dark fringe

We know that for dark fringe it is given that

[tex]dsin\Theta =n\lambda[/tex]

For first fringe n = 1

So [tex]1850\times 10^{-9}\times sin20^{\circ} =1\times \lambda[/tex]

[tex]\lambda =632.73nm[/tex]

Final answer:

The wavelength of the light from the He-Ne laser is 618 nm.

Explanation:

To determine the wavelength of the light from the He-Ne laser, we can use the equation for the first dark fringe in a single-slit diffraction pattern:

sin(theta) = m * (lambda) / w

Where theta is the angle of the first dark fringe, m is the order of the fringe (which is 1 for the first dark fringe), lambda is the wavelength of the light, and w is the width of the single slit.

Plugging in the given values:

sin(20.0∘) = 1 * (lambda) / 1850 nm

Simplifying and solving for lambda:

(lambda) = 1850 nm * sin(20.0∘)

(lambda) = 618 nm

Therefore, the wavelength of the light from the He-Ne laser is 618 nm.

Answer:

v = 1166.9 m / s

Explanation:

The equation for caloric energy is

      Q = m [tex]c_{e}[/tex] ΔT = m ce ([tex]T_{f}[/tex]-T₀)

Where m is the mass, ce is the specific heat and DT is the temperature variation

In this case the cannonball has kinetic energy

      Em = K = ½ m v²

They indicate that mechanical energy is completely transformed into heat

      Q = Em

      m [tex]c_{e}[/tex] ([tex]T_{f}[/tex] - To) = ½ m v²

      v = √ 2 [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-To)

Let's calculate

      v = √ (2 450 (1811-298))

      v = 1166.9 m / s

When the block started to move at  constant speed it starts to gain a  kinetic energy but the gravitational potential energy increases . Thus, option c is correct.

Kinetic energy of an object is the energy generated by virtue of its motion whereas potential energy is generated by virtue of its position in a gravitational filed.

Both kind of energy depends on the mass of the object.

Kinetic energy KE = 1/2 mv² where v is the velocity. Kinetic energy reaches its maximum at maximum speed. Potential energy p = Mgh, where g is acceleration due to gravity and h is the height.

Thus when the block moves in the incline  with constant velocity and its mass is also constant, no change will be there for kinetic energy but its potential energy will increase since gravitational pull increases. Thus option c is correct.

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To develop the problem it is necessary to apply the concepts related to the induced voltage and its definition according to the magnetic field and angular velocity.

By definition the induced voltage or electromotive force is given by

[tex]\epsilon = BAN\omega[/tex]

Where,

B = Magnetic Field

A = Cross-sectional area

N = Number of turns

[tex]\omega[/tex]= Angular velocity

For the given case in the problem we will look for the angular velocity,

[tex]\omega = \frac{\epsilon}{BAN}\\\omega = \frac{1}{(0.5*10^{-4})(0.01)(100)}\\\omega = 2*10^4rad/s \\\omega = 2*10^4rad/s (\frac{1rev}{2\pi rad})\\\omega = 3183.098rev/s[/tex]

Therefore the number of revolutions is 3183.1rev/s.

Although the number of revolutions the magnetic field of the earth is discarded for use as a generator because the magnetic field of the earth compared to other current devices is considered weak. Very little power would be obtained from the generator

Answer:

[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, [tex]\omega=2.8\ rad/s[/tex]

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

[tex]K_t=\dfrac{1}{2}mv^2[/tex]

[tex]K_t=\dfrac{1}{2}m(r\omega)^2[/tex]

[tex]K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2[/tex]

The rotational kinetic energy of the ball is :

[tex]K_r=\dfrac{1}{2}I \omega^2[/tex]

[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2[/tex]

[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2[/tex]

Ratio of translational to the rotational kinetic energy as :

[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.

Given the data in question;

Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}} = \ ?[/tex]

Rotational energy or angular kinetic energy is simply the kinetic energy due to the rotation of a rigid body while translational kinetic energy is the same as one-half the product of mass and the square of the velocity of the body.

They are expressed as;

[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]

Where;

Hence, [tex]E_{rotational} = \frac{1}{2}(\frac{2}{5}mR^2 )w^2[/tex]

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

Where;

Hence, [tex]E_{translational} = \frac{1}{2}m(R*w)^2[/tex]

To determine the translational kinetic energy of the bowling ball, we substitute our given values into the equation above.

[tex]E_{translational} = \frac{1}{2}m( R*w)^2\\ \\E_{translational} = \frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]

To determine the rotational kinetic energy of the bowling ball, we substitute our given values into the equation above.

[tex]E_{rotational} = \frac{1}{2}( \frac{2}{5}mR^2)w^2\\ \\E_{rotational} = \frac{1}{2}( \frac{2}{5} * 5.0kg* (0.11m)^2)(2.80rad/s)^2\\ \\E_{rotational} = \frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]

So, Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}[/tex]

[tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}{\frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} }{\frac{2}{10} }\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1*10}{2} }{2}\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{5}{2} = 2.50[/tex]

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.

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Final answer:

The magnitude of the induced emf in the coil as a function of time is given by the equation E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3. The current in the resistor at time t0 = 5.25 s can be calculated using Ohm's law by dividing the induced emf in the coil by the resistance.

Explanation:

In order to find the magnitude of the induced emf in the coil as a function of time, we can use Faraday's law of electromagnetic induction. Faraday's law states that the induced emf in a coil is equal to the rate of change of magnetic flux through the coil. The magnetic flux through the coil can be determined by multiplying the magnetic field by the area of the coil. Therefore, the induced emf in the coil is given by the equation E = -N * dΦ/dt, where E is the induced emf, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.

In this case, the magnetic field varies with time according to B = (1.20×10^(-2) T/s)t + (3.45×10^(-5) T/s^4)t^4. The magnetic field is perpendicular to the plane of the coil, so the area of the coil through which the magnetic flux passes is given by A = πr^2, where r is the radius of the coil.

Substituting these values into the equation for the induced emf, we get:

E = -N * (A * dB/dt)

E = -N * (πr^2 * dB/dt)

E = -520 * (π(0.0395)^2 * ((1.20×10^(-2) T/s) + (3.45×10^(-5) T/s^4)(4t^3)))

Simplifying this equation gives the magnitude of the induced emf in the coil as a function of time: E = 3.36×10^(-2) V + (3.30×10^(-4) V/s^3)t^3.

To find the current in the resistor at time t0 = 5.25 s, we can use Ohm's law, which states that the current through a resistor is equal to the voltage across the resistor divided by the resistance. The voltage across the resistor is equal to the induced emf in the coil. Therefore, the current in the resistor is given by the equation I = E/R, where I is the current, E is the induced emf, and R is the resistance.

Substituting the values given in the question into this equation, we get:

I = (3.36×10^(-2) V + (3.30×10^(-4) V/s^3)(5.25 s)^3) / 560 Ω

Calculating this gives the current in the resistor at time t0 = 5.25 s.

Answer:

The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

Explanation:

Given:

Mass of the driver is, [tex]m=55\ kg[/tex]

Radius of circular turn is, [tex]R=300\ m[/tex]

Linear speed of the car is, [tex]v=22.2\ m/s[/tex]

Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.

The forces are:

1. Weight = [tex]mg=55\times 9.8=539\ N[/tex]

2. Centripetal force, 'F', which is given as:

[tex]F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N[/tex]

Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,

[tex]\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10[/tex]°

Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

The net force angle relative to the vertical is found by calculating the centripetal and vertical forces. The angle is approximately 10°. This involves understanding forces in circular motion.

To find the angle of the net force of the car seat on the police officer, we need to consider the forces involved and the direction of the net force.

The car is executing horizontal circular motion, which means there is a centripetal force acting horizontally towards the center of the circle. This force is provided by the friction between the car tires and the road. The centripetal force Fc can be calculated by:

Fc = m * ac

where m is the mass of the officer and ac is the centripetal acceleration, which is given by:

ac = v2 / r

Given:

Radius of the turn, r = 300 m

Speed of the car, v = 22.2 m/s

Mass of the officer, m = 55.0 kg

First, calculate the centripetal acceleration:

ac = (22.2 m/s)2 / 300 m = 1.6444 m/s2

Now, calculate the centripetal force:

Fc = 55.0 kg * 1.6444 m/s2 = 90.442 N

The vertical force acting on the officer is her weight:

Fw = m * g = 55.0 kg * 9.8 m/s2 = 539 N

Now, to find the angle θ relative to the vertical, use the tangent function, as the net force's components are the horizontal centripetal force and the vertical weight:

θ = tan-1(Fc / Fw) = tan-1(90.442 / 539)

θ ≈ 9.5°

Thus, the angle of the net force relative to the vertical is approximately 10° (to the nearest degree).

Answer:

Explanation:

Expression for amplitude of forced damped oscillation is as follows

A = [tex]\frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }[/tex]

where

ω₀ =[tex]\sqrt{\frac{k}{m} }[/tex]

ω₀ = [tex]\sqrt{\frac{500000}{350} }[/tex]

=37.8

b = .125 ,

ω = 2.5

m = 350  

A = [tex]\frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }[/tex]

A = 3.75 mm . Ans

To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.

The two definitions of pressure can be enclosed under the following equations

[tex]P = \frac{F}{A}[/tex]

Where

F= Force

A = Area

[tex]P = \rho gh[/tex]

Where,

[tex]\rho =[/tex] Density

g = Gravity

h = Height

Our values are given as,

[tex]d = 0.8m \rightarrow r = 0.4m[/tex]

[tex]A = \pi r^2 = \pi * 0.4^2 = 0.503m^2[/tex]

If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.

Matching the two previous expressions we have to

[tex]\frac{F}{A} = \rho g h[/tex]

Re-arrange to find h

[tex]h = \frac{F}{A\rho g}[/tex]

[tex]h = \frac{390}{(0.503)(1000)(9.8)}[/tex]

[tex]h = 0.079m[/tex]

[tex]h = 7.9cm[/tex]

Therefore the height of water in the tube is 7.9cm

Answer:

6633549.52903 m

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.972 × 10²⁴ kg

[tex]h_p[/tex] = Height above ground = 227 km

[tex]v_p[/tex] = Velocity at perigee = 8.95 km/s

Perigee distance is

[tex]R_p=6371+227=6598\ km[/tex]

The apogee distance is given by

[tex]R_a=\dfrac{R_p}{\dfrac{2Gm}{R_pv_p^2}-1}\\\Rightarrow R_a=\dfrac{6598\times 10^3}{\dfrac{2\times 6.67\times 10^{-11}\times 5.972\times 10^{24}}{6598\times 10^3\times (8.950\times 10^3)^2}-1}\\\Rightarrow R_a=13004549.52903\ m[/tex]

The height above the ground would be

[tex]h_a=13004549.52903-6371000=6633549.52903\ m[/tex]

The height above the ground is 6633549.52903 m

Answer:

The velocities of hoop , disk and sphere are 1.89 m/s , 2.18 m/s , 2.26 m/s.

Explanation:

Lets find the speed of any general body of mass 'm' , moment of inertia 'I' , radius 'r'.

Let 'v' be the speed and 'ω' be the angular velocity of the body at  bottom of the slope.

Since there is no external force acting on the system (Eventhough friction is acting at the point of contact of the body and slope , it does no work as the point of contact is always at rest and not moving) , we can conserve energy for this system.

Initially the body is at rest and at a vertical height 'h' from the ground.

Here , h=3sin(7°)

Initial energy = mgh.

Finally on reaching bottom h=0 but the body has both rotational and translational kinetic energy.

∴ Final energy = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex].

Since the body is rolling without slipping.

v=rω

and

Initial Energy = Final Energy

mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]

∴ mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]\dfrac{v^{2} }{r^{2} }[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]

∴ v = [tex]\sqrt{\frac{2mgh}{\frac{I}{r^{2} }+m } }[/tex]

For a hoop ,

I = m[tex]r^{2}[/tex]

Substituting above value of I in the expression of v.

We get,

v = [tex]\sqrt{gh}[/tex] = [tex]\sqrt{9.81×3sin(7°) }[/tex] = 1.89 m/s

Similarly for disk,

I = [tex]\dfrac{1}{2}[/tex]m[tex]r^{2}[/tex]

We get,

v = [tex]\sqrt{\frac{4gh}{3} }[/tex] = 2.18 m/s

For solid sphere ,

I = [tex]\dfrac{2}{5}[/tex]m[tex]r^{2}[/tex]

v = [tex]\sqrt{\frac{10gh}{7} }[/tex] = 2.26 m/s.

The object with the greater mass should be attached to the spring with the smaller spring constant, so that the resulting spring-object system has the greatest possible period of oscillation.

Answer: Option D

Explanation:

According to the simple harmonic motions, from physics, it gives a relation between deformation force and the deflection. The more deflection results in more time period of oscillation.  

                                            F = - k x

where ‘k’ is the spring constant, and ‘F’ is the deformation force.

So, deflection is directly proportionate to forces, and inversely proportionate to its spring constant. Hence, we can derive that the force must be maximum, and hence weight must be maximum, with the spring constant lesser. Then, the deflection will be high. So, time period increases.

Answer:

a) The velocity of car B after the collision is 4.45 m/s.

The velocity of car A after the collision is 3.65 m/s.

b) The change of momentum of car A is - 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

Explanation:

Hi there!

Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.

The momentum of the system is calculated adding the momenta of each car:

initial momentum = final momentum

mA · vA + mB · vB = mA · vA´ + mB · vB´

Where:

mA = mass of car A

vA = initial velocity of car A

mB = mass of car B

vB = initial velocity of car B

vA´= final velocity of car A

vB´ = final velocity of car B

Let´s replace with the data we have and solve the equation for vA´:

mA · vA + mB · vB = mA · vA´ + mB · vB´

435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s - 495 kg · vB´ = 435 kg · vA´

(3789 kg · m/s - 495 kg · vB´)/435 kg = vA´

Let´s write this expression without units for a bit more clarity:

vA´= (3789 - 495 vB´)/435

The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:

initial kinetic energy of the system = final kinetic energy of the system

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA´)² + 1/2 · mB · (vB´)²

Replacing with the data:

initial kinetic energy = 1/2 · 435 kg · (4.50 m/s)² + 1/2 · 495 kg · (3.70)²

initial kinetic energy = 7792.65 kg · m²/s²

7792.65 kg · m²/s² = 1/2 · 435 kg · (vA´)² + 1/2 · 495 kg · (vB´)²

multiply by 2 both sides of the equation:

15585.3 kg · m²/s² =  435 kg · (vA´)² + 495 kg · (vB´)²

Let´s replace vA´ = (3789 - 495 vB´)/435

I will omit units for clarity in the calculation:

15585.3  =  435 · (vA´)² + 495 · (vB´)²

15585.3  =  435 · (3789 - 495 vB´)²/ 435² + 495 (vB´)²

15585.3 = (3789² - 3751110 vB´ + 245025 vB²) / 435 + 495 (vB´)²

multiply both sides of the equation by 435:

6779605.5 = 3789² - 3751110 vB´ + 245025 vB² + 215325 vB´²

0 = -6779605.5 + 3789² - 3751110 vB´ + 460350 vB´²

0 = 7576915.5 - 3751110 vB´ + 460350 vB´²

Solving the quadratic equation:

vB´ = 4.45 m/s

vB´ = 3.70 m/s (the initial velocity)

a) The velocity of car B after the collision is 4.45 m/s

The velocity of car A will be teh following:

vA´= (3789 - 495 vB´)/435

vA´= (3789 - 495 (4.45 m/s))/435

vA´ = 3.65 m/s

The velocity of car A after the collision is 3.65 m/s

b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:

ΔpA = final momentum of car A - initial momentum of car A

ΔpA = mA · vA´ - mA · vA

ΔpA = mA (vA´ - vA)

ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s)  (I have used the value of vA´ without rounding).

ΔpA = - 370.45 kg · m/s

The change of momentum of car A is - 370.45 kg · m/s

ΔpB = mB (vB´ - vB)

ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB´ without rounding).

ΔpB = 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.

In Physics, when discussing collisions, the total momentum of a closed and isolated system is conserved. The change in momentum during a collision can be calculated using the principles of conservation of momentum and energy, with specifics depending on whether the collision is elastic or inelastic.

The subject of this question is Physics, specifically dealing with the concepts of elastic and inelastic collisions, momentum, and the conservation of momentum principle. When two objects collide, momentum is exchanged between them, but the total momentum of the system remains constant if the system is closed and isolated. The change in momentum depends on whether the collision is elastic, where kinetic energy is conserved, or inelastic, where the objects stick together and kinetic energy is not conserved.

The example given with cars of equal mass indicates that after an elastic collision, the cars will trade velocities, and after a completely inelastic collision, they will move together with a combined momentum equal to their total initial momentum.

However, for the specific question asked by the student, which involves two bumper cars with different masses and velocities, the velocities after the collision and the change in momentum for each car would be determined by applying the elastic collision equations. These include conservation of momentum and conservation of kinetic energy. The precise calculations were not provided as part of the task.

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Answer:

Explanation:

Since the sled plus passenger moves with constant velocity , force applied will be equal to frictional force. Let the force applied be F

a ) Frictional force = μ R = F cosφ

R = mg - F sinφ

μ(mg - F sinφ)  = F cosφ

μmg = F (μsinφ+cosφ)

F = μmg / (μsinφ+cosφ)

Work done

= F cosφ x d

= μmg x cosφ x d / (μsinφ+cosφ)

b )Work done

= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / ( 0.13 sin36.7 +cos36.7)

= 1164.61 / .87946

1324.23 J

c ) work done on the sled by friction

= - (work done by force)

= - μmg x cosφ x d / (μsinφ+cosφ)

d ) work done on the sled by friction

= - 1324.23 J

The work done by the pulling force is given by the formula W = F * d * cos(φ), and in this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ).

The work done by the pulling force is given by the formula W = F * d * cos(φ), where F is the magnitude of the pulling force, d is the distance, and φ is the angle above the horizontal. In this case, the force of friction can be calculated as Ff = μk * m * g * cos(φ). Therefore, the work done by the pulling force can be written as W = Ff * d.

Using the given values, we can substitute in the values to calculate the work done by the pulling force, which is W = μk * m * g * cos(φ) * d.

The work done on the sled by friction can be calculated by multiplying the force of friction by the distance, since the force is acting in the opposite direction. Therefore, the work done on the sled by friction is given by Wfr = -Ff * d. Substituting the given values, we can calculate the work done by friction, which is Wfr = -μk * m * g * cos(φ) * d.

Inserting given values in this formula, we can calculate the work done by friction as Wf = - (0.13) * (52.3 kg) * (9.8 m/s²) * (21.8 m) = -1475.24 Joules.

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Answer:

The plane will be 11545.46 m far when the observer hears the sonic boom

Explanation:

Step 1: Data given

Altitude of the plane = 7800 meters

speed of sound = 311.83 m/s

Step 2:

The mach number M = vs/v

This means v/vs = 1/M

Half- angle = ∅

sin∅= v/vs

∅ = sin^-1 (v/vs)

∅ = sin^-1 (1/M)

∅ = sin^-1(1/1.48)

∅= 42.5 °

tan ∅ = h/x

⇒ with h= the altitude of the plane = 7800 meter

⇒ with x = the horizontal distance moved by the plane

x = h/tan ∅

x = 7800 / tan 42.5

x = 8512.2 meters

d = the distance between the observer and the plane when the observer hears the sonic boom is:

d = √(8512.2² + 7800²)

d = 11545.46 m

The plane will be 11545.46 m far when the observer hears the sonic boom

Explanation:

An object is attached to the spring and then released. It begins to oscillate. If it is displaced a distance 0.125 m from its equilibrium position and released with zero initial speed. The amplitude of a wave is the maximum displacement of the particle. So, its amplitude is 0.125 m.  

After 0.800 seconds, its displacement is found to be a distance 0.125 m on the opposite side. The time period will be, [tex]t=2\times 0.8=1.6\ s[/tex]

We know that the relation between the time period and the time period is given by :

[tex]f=\dfrac{1}{t}[/tex]

[tex]f=\dfrac{1}{1.6}[/tex]

f = 0.625 Hz

So, the frequency of the object is 0.625 Hz. Hence, this is the required solution.

Answer:

The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.

Explanation:

Given that,

Distance = 0.125 m

Time = 0.800 s

Since the object is released form rest, its initial displacement = maximum displacement .

(a). We need to calculate the amplitude

The amplitude is maximum displacement.

[tex]A=0.125 m[/tex]

(b). We need to calculate the period

The object will return to its original position after another 0.820 s,

So the time period will be

[tex]T=2\times t[/tex]

Put the value into the formula

[tex]T=2\times0.820[/tex]

[tex]T=1.64\ sec[/tex]

(c). We need to calculate the frequency

Using formula of frequency

[tex]f=\dfrac{1}{T}[/tex]

Put the value into the formula

[tex]f=\dfrac{1}{1.64}[/tex]

[tex]f=0.61\ Hz[/tex]

Hence, The amplitude is 0.125 m, period is 1.64 sec and frequency is 0.61 Hz.

This is a problem based on the logic and interpretation of the variables. From the measured data taken

what is collected by the two individuals is expressed as,

- NED reference system: (x, t)

- PAM reference system: (x ', t')

From the reference system we know that ν is the speed of PAM (the other reference system) as a measurement by NED.

Then ν' is the speed of NED (from the other system of the reference) as a measurement by PAM.

In the context of Special Relativity, v' represents the relative velocity of Ned as observed from Pam's reference frame. This velocity takes into account the motion of both Ned and Pam and can be computed using the principles of Conservation of Momentum.

The symbol v' can be interpreted as the velocity of Ned as measured from Pam's frame of reference. This concept forms the basis of Special Relativity as proposed by Einstein, where observations are taken from different frames of reference, in this case, Pam and Ned. Relative velocity, like v', is the velocity of an object (in this case, Ned) as observed from a particular frame of reference (in this case, Pam's), and it can change depending on who is making the observation. It takes into account the motion of both Ned and Pam and describes how fast Ned is moving and in what direction from Pam's perspective.

For example, if Pam and Ned were moving towards each other, the relative velocity v' would be the sum of their individual speeds as observed from Pam's perspective. If they were moving in the same direction, v' would be the difference in their individual speeds. Also, Pam's velocity in Ned's frame can be defined as -v'. This negative sign indicates they are moving in opposite directions.

To compute v', you would need to understand the principle of Conservation of Momentum, represented by equations defining momentum conservation in the x and y directions. Using this principle, you can solve for V.

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Answer:

The rate of heat rejection will be 347 GJ/hr

Explanation:

We have given that power plant uses energy of 365 GJ/hr

So energy uses [tex]Q_B=365GJ/hr[/tex]

Work done from turbine [tex]W_T=18GJ/hr[/tex]

We have to find the rate of heat rejection to the atmosphere [tex]Q_C[/tex]

We know that [tex]Q_B=W_T+Q_C[/tex]

[tex]365=18+Q_C[/tex]

[tex]Q_C=347GJ/hr[/tex]

So the rate of heat rejection will be 347 GJ/hr

Answer:

[tex]m_1( g -a) = T_1[/tex]

[tex]T_2 = m_2 (a +g)[/tex]

Explanation:

Given data;

Two hanging mass is given as m1 and m2

Mass of pulley is given as m3

radius of pulley is r

Assuming mass m1 is greater than m2

Take downward direction for mass m1

and upward direction for mass m2

and clockwise direction of pulley is positive

from newton second law on each mass

for Mass m1

[tex]\sum F_y = m_1 a_y[/tex]

[tex]m_1 g -T_1 = m_1 a[/tex]

[tex]m_1( g -a) = T_1[/tex]

for Mass m2

[tex]\sum F_y = m_2 a_y [/tex]

[tex]T_2 -  m_2 g = m_2 a[/tex]

[tex]T_2 = m_2 (a +g)[/tex]

fro pulley

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

[tex]rT_1 -rT_2 =1/2 m_p r^2 \alpha[/tex]