The Balmer series is formed by electron transitions in hydrogen that
a.begin on the n = 1 shell.
b. end on the n = 2 shell.
c. end on the n = 1 shell.
d. are between the n = 1 and n = 3 shells.
e. begin on the n = 2 shell.

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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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

The calculated tension is in the cable at the end of the scaffolding is 208.94 N.

First, let's calculate the weights acting on the system: the weight of the watermelon (Wm) and the weight of the scaffolding (Ws). Wm = mass of the watermelon × gravity = 7.1 kg × 9.8 m/s2 = 69.58 N, and Ws = 260 N given. Next, we choose the pivot point at the end of the scaffolding opposite the watermelon to find the tension in the cable at that end. The distance from the watermelon to the supporting cable is (4.8 m - 0.53 m) = 4.27 m, and the distance from the pivot to the scaffold's center of mass (assuming it's uniform) is half its length, or 2.4 m.

Applying the equilibrium condition for torques (Τclockwise = Τcounterclockwise), we have:

(69.58 N × 4.27 m) + (260 N × 2.4 m) = T × 4.8 m

⇒ T = ((69.58 N × 4.27 m) + (260 N × 2.4 m)) / 4.8 m

⇒ T = 208.94 N.

The tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].

Given:

- Mass of the watermelon, [tex]\( m = 7.1 \)[/tex] kg

- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s²

- Length of the scaffolding, [tex]\( L = 4.8 \)[/tex] m

- Distance of the watermelon from one end, [tex]\( d = 0.53 \)[/tex] m

1. Calculate the weight of the watermelon:

[tex]\[ W = mg = 7.1 \times 9.8 = 69.58 \text{ N} \][/tex]

2. Determine the tensions in the cables:

 Since the system is in equilibrium, the tension [tex]\( T_1 \)[/tex] at the end where the watermelon is placed and the tension [tex]\( T_2 \)[/tex] at the opposite end satisfy:

[tex]\[ T_1 + T_2 = W \][/tex]

3. Find [tex]\( T_2 \)[/tex]:

From the equilibrium condition:

[tex]\[ T_2 = \frac{W}{2} = \frac{69.58}{2} = 34.79 \text{ N} \][/tex]

4. Calculate [tex]\( T_1 \)[/tex]:

  Since [tex]\( T_1 + T_2 = W \)[/tex] and [tex]\( T_1 = -T_2 \)[/tex] (because [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] act in opposite directions):

[tex]\[ T_1 = W - T_2 = 69.58 - 34.79 = 34.79 \text{ N} \][/tex]

Therefore, the tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].

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:

S = 2266.67 m/s

Explanation:

Given,

length of the metal = 3.4 m

pulses are separated in time = 8.4 ms

speed of sound in air= 343 m/s

speed of sound in this metal = ?

time taken

[tex]t = \dfrac{distance}{speed}[/tex]

[tex]t = \dfrac{3.4}{343}[/tex]

t = 9.9 ms

speed of sound in the metal is fast

t = 9.9 - 8.4 = 1.5 ms

time for which sound is in metal is equal to 1.5 ms

speed of sound in metal

 [tex]speed= \dfrac{distance}{time}[/tex]

[tex]S = \dfrac{3.4}{1.5 \times 10^{-3}}[/tex]

S = 2266.67 m/s

Speed of sound in metal is equal to S = 2266.67 m/s

Final answer:

The speed of sound in the metal is approximately 404.76 m/s.

Explanation:

To determine the speed of sound in the metal, we need to use the time difference between the two pulses and the length of the metal bar. The time difference between the pulses is 8.40 ms and the length of the bar is 3.4 m. The speed of sound in the metal can be calculated using the formula: speed = distance / time. In this case, the distance is the length of the bar and the time is the time difference between the pulses.

So, by substituting the values into the formula, we get: speed = 3.4 m / (8.40 * 10^-3 s).

Simplifying the equation, we get: speed = 404.76 m/s.

Therefore, the speed of sound in the metal is approximately 404.76 m/s.

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

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

To solve this problem, it is necessary to apply the concepts of the Simple Pendulum Period. Under this definition it is understood as the time it takes for the pendulum to pass through a point in the same direction. It is also defined as the time it takes to get a complete swing. Its value is determined by:

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

Where,

T= Period

l = Length

g = Gravitaitonal Acceleration

With our values we have tat

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

[tex]T = 2\pi \sqrt{\frac{67}{9.809}}[/tex]

[tex]T = 16.413s[/tex]

Therefore the period of the pendulum is 16.4s

The period of a pendulum is calculated using the formula T = 2π √(L/g), where T is the period, L is the length and g is the acceleration due to gravity. In the given scenario, values are provided for L and g, which can be substituted into the formula to find the period. Importantly, the mass of the pendulum does not affect the period.

The period of a pendulum can be calculated using the formula T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. In the given problem, the length, L, of the pendulum is 67.00 m and the acceleration due to gravity, g, in Paris is 9.809 m/s². Using these values in the formula, the period T would be T = 2π √(67.00 m / 9.809 m/s²).

It's important to note, as shown in the formula, that the period of a pendulum depends solely on its length and the acceleration due to gravity. The mass of the pendulum does not affect the period. This is one of the unique properties of a pendulum and is the principle behind its use in clocks and other timing devices.

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Answer:

The no. of capillaries are [tex]1.92\times 10^{10}[/tex]

Solution:

As per the question:

Speed of the blood carried by the aorta, [tex]v_{a} = 40\ cm/s[/tex]

Radius of the aorta, [tex]R_{a} = 1.1 cm[/tex]

Speed of the blood in the capillaries, [tex]v_{c} = 0.007\ cm/s[/tex]

Radius of the capillaries, [tex]R_{c} = 6.0\times 10^{- 4} cm[/tex]

Now,

To determine the no. of capillaries:

Cross sectional Area of the Aorta, [tex]A_{a} = \pi R_{a}^{2} = \pi \times (1.1)^{2} = 1.21\pi \ m^{2}[/tex]

Cross sectional Area of the Capillary, [tex]A_{c} = \pi R_{c}^{2} = \pi \times (6.0\times 10^{- 4})^{2} = (3.6\times 10^{- 7})\pi \ m^{2}[/tex]

Let the no. of capillaries be 'n'

Also, the volume rate of flow in the aorta equals the sum total flow in the 'n' capillaries:

[tex]A_{a}v_{a} = nA_{c}v_{c}[/tex]

[tex]1.21\pi\times 40 = n\times 3.6\times 10^{- 7}\pi\times 0.007[\tex]

[tex]n = 1.92\times 10^{10}[/tex]

Using the principle of continuity, which states the constant volume flow rate of an incompressible fluid, we calculate that the human body has approximately ten billion capillaries.

To determine the approximate number of capillaries in the human body based on the given data, we can use the principle of continuity, which states that the product of the cross-sectional area of a tube and the fluid speed through the tube is constant.

This implies that the blood flow (volume rate of flow) is the same in the aorta as in the capillaries, i.e., Aorta's cross-sectional area × speed of blood flow in Aorta = Capillary's cross-sectional area × speed of blood flow in capillaries × number of capillaries. From this, we can solve for the number of capillaries:

(π(1.1 cm)^2 × 40 cm/s) = (π(6.0 × 10^-4 cm)^2 × 0.007 cm/s) × number of capillaries.

When we do the calculations, we find that there are approximately ten billion capillaries in the human body, a vast network to ensure blood is delivered to every part of the body.

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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.

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:

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

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

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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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.

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:

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:

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:

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:

[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) 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: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 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.

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:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

[tex]V_e=\sqrt{\frac{2GM}{R}}[/tex]

Escape velocity at the surface of earth

[tex]11=\sqrt{\frac{2GM}{R}}[/tex]--------------------1

If Escape velocity is three times and the radius is also the three times

[tex]V_e_2=\sqrt{\frac{2G(3M)}{3R}}[/tex]

[tex]V_e_2=\sqrt{\frac{2GM}{R}}=V_e[/tex]

i.e. [tex]V_e_2=11 km/s[/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:

[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]

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.