A 52 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. The acceleration of gravity is 9.8 m/s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.

Answer:

120.9 seconds

Explanation:

*Attached below is a rough sketch of the problem. Point A represents the point where the airplane climbs 30°, BC represents the altitude of the airplane, AC represents the angular displacement of the airplane.

Parameters given:

Angle of elevation = 30°

Speed (angular) of the airplane = 215 ft/sec

Altitude = 13000 ft

We need to find the angular displacement of the airplane to find the time it takes to get to an altitude of 13000 ft. The angular displacement is represented by the hypotenuse of the triangle. Hence, using SOHCAHTOA,

[tex]sin30^{o} = \frac{13000}{hyp} \\\\hyp = \frac{13000}{sin30^{o}}\\ \\hyp = 26000 ft[/tex]

∴ time taken = [tex]\frac{angular displacement}{angular speed}\\ \\[/tex]

[tex]time taken = \frac{26000}{215} \\\\time taken = 120.9 seconds[/tex]

Therefore, it will take the airplane 120.9 seconds to get to an altitude of 13000 ft.

Answer:

Distance = 26.0m Displacement = 4.0m

Explanation:

Distance specifies only how far an object has traveled while displacement is the distance traveled in a specified direction.

Total distance traveled by the object will be distance travelled through north + distance travelled through south i.e 15.0m + 11.0m = 26.0m

Displacement is gotten by using the Pythagoras theorem. Since the object traveled in the same vertical direction (15.0m through north which is upward i.e positive y direction and 11.0m through south i.e in the negative y direction), the displacement will be 15.0m - 11.0m = 4.0m

The distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

First, we will define the terms distance and displacement

Distance is the total movement of an object without any regard to direction.

Displacement is the difference between the original and final position of a path taken by an object.

Since, the object moves 15.0 m north and then 11.0 m south,

Then,

Distance traveled = 15.0 m + 11.0 m

Distance traveled = 26.0 m

For the magnitude of the displacement,

The object moves 15.0 m north and then 11.0 m south, which is in the opposite (negative) direction

Then,

Magnitude of displacement = 15.0 m - 11.0 m

Magnitude of displacement = 4.0 m

Hence, the distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

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Final answer:

The sound intensity of the full orchestra is ten times greater than that of the violin section alone because a 10 dB increase corresponds to a tenfold increase in intensity.

Explanation:

The decibel level (dB) measures sound intensity on a logarithmic scale, where each 10 dB increase represents a tenfold increase in intensity. To compare the sound intensity of the full orchestra at 90 dB to the violin section at 80 dB, we can use the fact that an increase of 10 dB corresponds to a tenfold increase in intensity. Therefore, the full orchestra's sound intensity is ten times greater than that of the violin section alone.

This simplifies to 10, meaning that the sound intensity from the full orchestra is 10 times greater than that from the violin section alone.

In simpler terms, when comparing sound intensities, every 10 dB increase corresponds to a tenfold increase in intensity. Therefore, the difference of 10 dB between the full orchestra and the violin section results in the orchestra being 10 times louder in terms of sound intensity.

The sound intensity from the full orchestra is [tex]\( {10} \)[/tex] times greater than the sound intensity from the violin section alone.

To compare the sound intensities of the full orchestra and the violin section, we need to understand the relationship between decibels (dB) and sound intensity [tex]\( I \)[/tex].

The decibel scale is logarithmic and relates to sound intensity [tex]\( I \)[/tex] in the following way:

[tex]\[ \text{dB} = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]

where [tex]\( I \)[/tex] is the sound intensity of interest and [tex]\( I_0 \)[/tex] is the reference intensity (typically [tex]\( I_0 = 10^{-12} \)[/tex] W/m[tex]\(^2\))[/tex].

Step 1: Convert dB to intensity ratio

Given:

- Orchestra sound level [tex]\( L_{\text{orchestra}} = 90 \)[/tex] dB

- Violin section sound level [tex]\( L_{\text{violin}} = 80 \)[/tex] dB

The difference in decibel levels between the orchestra and the violin section gives us the intensity ratio:

[tex]\[ L_{\text{orchestra}} - L_{\text{violin}} = 90 \, \text{dB} - 80 \, \text{dB} = 10 \, \text{dB} \][/tex]

The intensity ratio in terms of decibels is related by:

[tex]\[ 10 \log_{10} \left( \frac{I_{\text{orchestra}}}{I_0} \right) - 10 \log_{10} \left( \frac{I_{\text{violin}}}{I_0} \right) = 10 \][/tex]

Step 2: Calculate the ratio of sound intensities

To find [tex]\( \frac{I_{\text{orchestra}}}{I_{\text{violin}}} \)[/tex]:

[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{10 / 10} \][/tex]

[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10^{1} \][/tex]

[tex]\[ \frac{I_{\text{orchestra}}}{I_{\text{violin}}} = 10 \][/tex]

The primary current for the given transformer is 3 A which is determined by considering the principle of a transformer, the turns ratio and the law of conservation of power, and calculating secondary current using Ohm's law.

To find the primary current of a transformer, we need to first understand the principle of a transformer. A transformer operates on the electromagnetic induction principle, and the ratio of turns in the primary and secondary coil determines how much voltage is changed. In your transformer, the primary voltage is 240 V and the secondary voltage is 60 V. Therefore, the turns ratio is 240/60 which gives a value of 4.

Now, considering the law of conservation of power in an ideal transformer, the power output is equal to the power input. Power is given by the product of voltage and current.

So, primary voltage (V1) times primary current (I1) is equal to secondary voltage (V2) times secondary current (I2). The secondary current (I2) can be calculated using Ohm's law as V2/R, where R is the resistance of the load connected. Plugging in the given values we get

I2 = 60/5 = 12 A.

Now, substituting the known variables in the power conservation equation we get

240*I1 = 60*12.

Solving for I1 gives a value of 3A.

Therefore, the primary current for the given transformer is 3 A.

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Answer:

Δβ = 28.2 dB

Explanation:

Attached is the explanation

Answer:

-387883.3 m/s²

0.000286168546055 seconds

Explanation:

t = Time taken

u = Initial velocity = 370 m/s

v = Final velocity = 259 m/s

s = Displacement = 9 cm

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2[/tex]

The acceleration of the bullet is -387883.3 m/s²

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s[/tex]

The time taken is 0.000286168546055 seconds

Final answer:

The acceleration of the bullet is approximately -111 m/s^2 and the total time the bullet is in contact with the board is approximately 0.00081 seconds.

Explanation:

The acceleration of the bullet can be calculated using the equation of motion:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the initial velocity of the bullet is 370 m/s (positive because it is in the direction of motion) and the final velocity is 259 m/s (positive because it is also in the direction of motion). The time it takes for the bullet to decelerate can be calculated using the equation:

x = ut + (1/2)at^2

where x is the displacement, u is the initial velocity, t is the time, and a is the acceleration. In this case, the displacement is the thickness of the wooden board, which is 9.0 cm (or 0.09 m). Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we have:

t = (259 - 370) / a

Combining the equations for time and acceleration, we get:

t = (0.09) / (370 - 259) = (0.09) / (111) = 0.00081 seconds.

So, the acceleration of the bullet is approximately -111 m/s^2 (negative because it is in the opposite direction of the bullet's motion) and the total time the bullet is in contact with the board is approximately 0.00081 seconds.

Answer:

The concepts of basic chemistry and physics teach us that the more stable a compound would be, the more difficult it will be to break that compound. On the other hand, compounds which are not stable have weak binds and can be  broken down easily.

For example, CFC's are stable compounds. It is difficult to break down these compounds and hence they exist in the environment without being broken down. They cause harmful effects on the ozone layer of the Earth.

The question combines the principles of wave mechanics, specifically sinusoidal waves, to calculate the height of a resultant wave at a certain time and position. This is done by applying values to the formula y (x, t) = A sin (kx - wt +φ), where A, k, w and φ stand for amplitude, wave number, angular frequency and phase shift respectively. Applying the values of the moving sinusoidal waves, you can solve for the height of the resultant wave.

A sinusoidal wave is represented by the equation y (x, t) = A sin (kx — wt + φ). Here A is the amplitude, k is wave number, w is the angular frequency and φ is the phase shift. Given two sinusoidal waves, moving through a medium, both having the equal amplitude (7.00 cm), wave number (k=3.00m−1) and angular frequency (ω=2.50s−1), but with one having a phase shift of π/12 radians, the resultant wave has an amplitude of AR = [2A cos(φ/2)] and a phase shift equal to half the original phase shift.

Applying these values to the equation y (x, t) = AR sin (kx - wt + φ/2), we have y = 2*7* cos(π/12/2) cm * sin(3.00m−1 *0.53m — 2.50s−1* 2.00s + π/12 /2), thus we can calculate the height of the resultant wave at a particular position and time.

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The height of the resultant wave at [tex]\( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \)[/tex] is:

[tex]\[ \boxed{-13.965 \, \text{cm}} \][/tex].

The height of the resultant wave at time [tex]\( t = 2.00 \, \text{s} \)[/tex] and position [tex]\( x = 0.53 \, \text{m} \)[/tex] is given by the sum of the two individual waves, taking into account their respective phase shifts.

The general equation for a sinusoidal wave moving in the positive x-direction is:

[tex]\[ y(x,t) = A \cos(kx - \omega t + \phi) \][/tex]

[tex]where \( A \) is the amplitude, \( k \) is the wave number, \( \omega \) is the angular frequency, \( \phi \) is the phase shift, \( x \) is the position, and \( t \) is the time.[/tex]

For the first wave with no phase shift, the equation is:

[tex]\[ y_1(x,t) = 7.00 \cos(3.00x - 2.50t) \][/tex]

For the second wave with a phase shift [tex]\( \phi = \frac{\pi}{12} \)[/tex], the equation is:

[tex]\[ y_2(x,t) = 7.00 \cos(3.00x - 2.50t + \frac{\pi}{12}) \]\\ At \( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \), we substitute these values into the equations for \( y_1 \) and \( y_2 \): \[ y_1(0.53,2.00) = 7.00 \cos(3.00 \cdot 0.53 - 2.50 \cdot 2.00) \] \[ y_1(0.53,2.00) = 7.00 \cos(1.59 - 5.00) \] \[ y_1(0.53,2.00) = 7.00 \cos(-3.41) \][/tex]

Since the cosine function is even, [tex]\( \cos(-3.41) = \cos(3.41) \)[/tex], we can write:

[tex]\[ y_1(0.53,2.00) = 7.00 \cos(3.41) \][/tex]

For the second wave:

[tex]\[ y_2(0.53,2.00) = 7.00 \cos(3.00 \cdot 0.53 - 2.50 \cdot 2.00 + \frac{\pi}{12}) \] \[ y_2(0.53,2.00) = 7.00 \cos(1.59 - 5.00 + \frac{\pi}{12}) \] \[ y_2(0.53,2.00) = 7.00 \cos(-3.41 + \frac{\pi}{12}) \] The resultant wave \( y \) is the sum of \( y_1 \) and \( y_2 \): \[ y(0.53,2.00) = y_1(0.53,2.00) + y_2(0.53,2.00) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + \frac{\pi}{12}) \][/tex]

To find the numerical value, we calculate the cosines:

[tex]\[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + \frac{\pi}{12}) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.41 + 0.2618) \] \[ y(0.53,2.00) = 7.00 \cos(3.41) + 7.00 \cos(-3.1482) \][/tex]

Using a calculator or trigonometric tables, we find:

[tex]\[ y(0.53,2.00) = 7.00 \cdot (-0.9956) + 7.00 \cdot (-0.9994) \] \[ y(0.53,2.00) = -6.9692 - 6.9958 \] \[ y(0.53,2.00) = -13.965 \][/tex]

Therefore, the height of the resultant wave at [tex]\( t = 2.00 \, \text{s} \) and \( x = 0.53 \, \text{m} \)[/tex] is:

[tex]\[ \boxed{-13.965 \, \text{cm}} \][/tex]

The negative sign indicates that the displacement is in the negative y-direction.

Answer:

Approximately 500kJ

Explanation:

According to the law of conservation of energy which states that energy can neither be created nor destroyed but can be converted from one form to another.

A parachutist jumping out of the airplane covering a particular height and under the influence of gravity will possess potential energy during fall. Since PE = mass × acceleration due to gravity × height

PE = 39.4×1340×9.8

PE = 501,562Joules

If the body lands on the ground with a speed of 5.78m/s, this means the body possesses kinetic energy at the point of landing. The kinetic energy on landing is 1/2mv²

KE = 1/2×39.4×5.78²

KE = 658.14Joules

The amount of energy lost due to friction will be PE-KE

= 501,562-658.14

= 500,903Joules approximately 500kJ

Final answer:

The solution involves calculating the parachutist's initial potential energy and final kinetic energy, then finding the difference to determine the energy lost to air friction during the jump.

Explanation:

The question asks how much energy was lost to air friction during the jump of a parachutist who has a mass of 39.4 kg, jumps from a height of 1340 m, and lands with a speed of 5.78 m/s. To find the energy lost to air friction, we first calculate the potential energy at the start and the kinetic energy at the end of the jump, then find the difference.

Initial potential energy (PE_initial) is given by mgh, where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height. Therefore, PE_initial = 39.4 kg * 9.8 m/s² * 1340 m.

Final kinetic energy (KE_final) is given by 1/2 mv², where m is mass and v is velocity at landing. Thus, KE_final = 1/2 * 39.4 kg * (5.78 m/s)².

The energy lost to air friction equals the initial potential energy minus the final kinetic energy. Subtracting KE_final from PE_initial gives the amount of energy lost to air resistance.

Answer:

Explanation:

Given

Speed of ball [tex]u=3.2\ m/s[/tex]

Plane is inclined at an angle [tex]20^{\circ}[/tex]

To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away

Launch angle of ball [tex]\theta [/tex]

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is [tex]g\sin 20[/tex]

Range of Projectile is given by

[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]

for [tex]R=2.4\ m[/tex]

[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]

[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]

[tex]\sin 2\theta =0.7855[/tex]

[tex]2\theta =51.77[/tex]

[tex]\theta =25.88^{\circ}[/tex]

so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]

To answer this, we use projectile motion principles with an equation specifically structured for the problem. Inserting the given values into the equation derived from the horizontal and vertical equations of motion, we can find the required launching angle to hit the target.

To solve this problem, we can apply the concepts found in projectile motion physics. Given the initial speed or velocity (3.2 m/s) and the horizontal distance of the target (2.4 m) we can find the necessary angle (theta) to hit the target. The angle θ can be given by the equation of motion for a projectile which can be derived from the horizontal and vertical equations of motion, θ = atan[(vf² ± sqrt(vf⁴ - g*(g*x² + 2*y*vf²)) / (g*x)] where g = 9.8 m/s² is the acceleration due to gravity, x = 2.4 m is the horizontal distance, y = 0 (height difference), and vf = 3.2 m/s is the final velocity, the speed at which the ball is launched.

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Answer:

What is the coefficient of kinetic friction = 0.432

Explanation:

The detailed steps and derivation with appropriate substitution is as shown in the attached file.

This question involves the concepts of the law of conservation of energy and frictional energy.

a) The speed of the child when he reaches the bottom of the slide is "6 m/s".

b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".

a)

According to the law of conservation of energy in this situation:

Loss in Potential Energy = Gain in Kinetic Energy

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

where,

v = velocity = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = 6 ft = 1.83 m

Therefore,

[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]

v = 6 m/s

b)

Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:

Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy

[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]

where,

l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]

[tex]\mu_k[/tex] = coefficient of kinetic friction = ?

Therefore,

[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]

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The attached picture explains the law of conservation of energy.

Answer: The answer is attached

Explanation:

Final answer:

In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.

Explanation:

P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.

P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.

Answer : The energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J

Explanation :

As we are given that:

1 watt = 1 J/s

So,

100 watt = 100 J/s

Now we have to calculate the energy consumed by bulb in 24 hours.

As we know that:

1 hr = 60 min

1 min= 60 sec

So,

24 hr = 24 × 60 × 60 sec = 86400 sec

As, the energy consumed by bulb in 1 second = 100 J

So, the energy consumed by bulb in 86400 second = 86400 × 100 J

                                                                                       = 8.64 × 10⁶ J

Thus, the energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J

To calculate the energy consumed by a 100-watt electric incandescent light bulb in 24 hours, we use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s). In this case, the energy consumed is 8640000 J.

To calculate the electrical energy used by a 100-watt electric incandescent light bulb in 24 hours, we can use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s).

First, we need to convert the time from hours to seconds. There are 3600 seconds in an hour. So, 24 hours is equal to 24 x 3600 = 86400 seconds.

Now, we can substitute the values into the formula. E = 100W x 86400s = 8640000 J.

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Final answer:

The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.

Explanation:

a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.

b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.

c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.

Answer:

A) An interference pattern

Explanation:

the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.

Key Points

Answer: did you get the answer(s) by chance?

Explanation:

Answer:

The last one

Explanation:

Answer:

2.57 m/s

Explanation:

mass of Erica (ME) = 39 kg

velocity of Erica (VE) = 0 m/s (since she is at her high point of her jump, her velocity will be 0)

mass of Danny (MD) = 45 kg

velocity of Danny (VD) = 4.8 m/s

from the conservation of momentum total initial momentum is equal to the total final momentum

total initial momentum = final initial momentum

(ME)(VE) + (MD)(VD) = (ME + MD) V

where "V" is the velocity of Erica and Danny after Danny grabs her.

(39 x 0) + (45 x 4.8) = (39 + 45) x V

0 + 216 = 84 V

V = 216 / 84 = 2.57 m/s

The speed just after Danny grabs Erica is equal to 2.57 m/s.

Given the following data:

To determine their speed just after Danny grabs Erica, we would apply the law of conservation of momentum:

[tex]M_EV_E + M_DV_D = (M_E + M_D)V_{f}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]39 \times 0 + 45 \times 4.8 = (39 +45)V_f\\\\0+216=84V_f\\\\V_f = \frac{216}{84} \\\\V_f = 2.57 \;m/s[/tex]

Final speed = 2.57 m/s

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Answer:

the new volume =93,750 Liters

Explanation:

From Boyles law

P1V1 = P2V2

1.2 X 125000 = 1.6 X V2

V2 = 93,750 Liters

Answer:

The angle of the incident ray at the air-glass interface is 62.86°

Explanation:

Given that,

Refractive index of glass =1.60

Angle = 42°

We need to calculate the angle of the incident ray at glass-water interface

Using Snell's law

[tex]n_{2}\sin\theta_{2}=n_{3}\sin\theta_{3}[/tex]

Where, [tex]n_{2}[/tex] = refractive index of glass

[tex]n_{3}[/tex] = refractive index of water

[tex]\theta_{3}[/tex] = angle of refraction

Put the value into the formula

[tex]1.6\sin\theta_{2}=1.33\sin42[/tex]

[tex]\sin\theta_{2}=\dfrac{1.33\sin42}{1.6}[/tex]

[tex]\theta_{2}=\sin^{-1}(\dfrac{1.33\sin42}{1.6})[/tex]

[tex]\theta_{2}=33.8^{\circ}[/tex]

We need to calculate the angle of the incident ray at the air-glass interface

Using Snell's law

[tex]n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}[/tex]

Where, [tex]n_{1}[/tex] = refractive index of air

[tex]n_{2}[/tex] = refractive index of glass

[tex]\theta_{1}[/tex] = angle of incident

Put the value into the formula

[tex]1\sin\theta_{1}=1.6\sin33.8[/tex]

[tex]\theta=\sin^{-1}(1.6\sin33.8)[/tex]

[tex]\theta=62.86^{\circ}[/tex]

Hence, The angle of the incident ray at the air-glass interface is 62.86°

Answer:

62.8 degree

Explanation:

Let the incident ray incident at an angle [tex]\theta_1[/tex] at air glass surface.

[tex]\theta_3=42^{\circ}[/tex]=Angle of refraction when ray travel from glass to water

[tex]\theta_2=[/tex]Angle of refraction when the ray travel from air to glass

Refractive index of glass,[tex]n_2=1.6[/tex]

We know that

Refractive index of water=[tex]n_3=1.33[/tex]

Snell's law

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

Where [tex]\theta_1[/tex]=Angle of incidence

[tex]\theta_2=[/tex]Angle of refraction

[tex]n_1=[/tex]Refractive index of medium 1

[tex]n_2=[/tex]Refractive index of medium 2

When the ray travel from glass to water

[tex]n_2sin\theta_2=n_3sin\theta_3[/tex]

Where [tex]n_2=[/tex]Refractive index of medium 1(Glass)

[tex]n_3[/tex]=Refractive index of medium 2 (Water)

[tex]\theta_2=[/tex]Angle of incidence

[tex]\theta_3[/tex]=Angle of refraction

Substitute the values

[tex]1.6sin\theta_2=1.33sin42[/tex]

[tex]sin\theta_2=\frac{1.33sin42}{1.6}[/tex]

[tex]sin\theta_2=0.556[/tex]

[tex]\theta_2=sin^{-1}(0.556)=33.8^{\circ}[/tex]

Angle of refraction when ray travel from air to glass= Angle of incidence of when ray travel from glass to water

Angle of refraction when the ray travel from air to glass=33.8 degree

Refractive index of air=[tex]n_1=1[/tex]

Again apply Snell's law

[tex]n_1sin\thet_1=n_2sin\theta_2[/tex]

[tex]1\times sin\theta_1=1.6sin(33.8)[/tex]

[tex]sin\theta_1=1.6\times 0.556=0.8896[/tex]

[tex]\theta_1=sin^{-1}(0.8896)=62.8^{\circ}[/tex]

Hence, the angle of the incident ray at the air-glass interface=62.8 degree

Kepler's laws describe the motion of planets based on empirical observations without explaining why they behave in such a way. Newton's laws, particularly the law of universal gravitation, provide the reasons for these behaviors, explaining them using cause-effect relationships.

The fundamental differences between Kepler's laws and Newton's laws pertain to their descriptions of planetary motion. Kepler's laws are based on the empirical observations and provide a descriptive perspective on the motion of planets around the Sun in an elliptical orbit. They consist of three essential laws: the Law of Orbits, Law of Areas, and Law of Periods.

Newton's laws, on the other hand, provide an explanatory perspective grounded in cause-effect relationships. These laws, especially the law of universal gravitation, explain why the planets follow an elliptical orbit as Kepler observed, based on the gravitational influence of the Sun.

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The differences between Kepler's laws and Newton's laws are that Kepler described planetary motion through empirical observations, while Newton explained it theoretically with universal laws applicable to all motions in the universe.

The true statements regarding the differences between Kepler's laws and Newton's laws are:

A. Kepler reported how planets moved, and Newton explained why.

Kepler's laws describe the observed motions of planets, while Newton's laws provide a theoretical explanation for these motions based on the law of universal gravitation.

B. Kepler defined laws based on one special case - the observed motions of planets in the Solar System, while Newton defined theoretical laws that describe the general behavior of all motions in the universe.

Kepler's laws were specific to planetary motion in the Solar System, while Newton's laws of motion and the law of universal gravitation are general laws that apply to all objects and motions in the universe.

D. Kepler's approach was empirical, while Newton's was theoretical.

Kepler's laws were derived from observations, making his approach empirical. In contrast, Newton's laws were based on theoretical principles, including the law of universal gravitation.

To know more about solar system:

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The question is incomplete, the given complete question is:

"Both Kepler's laws and Newton's laws tell us something about the motion of the planets, but there are fundamental differences between them. What are the differences? Select all of the true statements. Choose one or more: O A. Kepler reported how planets moved, and Newton explained why. B. Kepler defined laws based on one special case-the observed motions of planets in the Solar System, while Newton defined theoretical laws that describe the general behavior of all motions in the universe O C. Kepler's discoveries built on the work of an earlier astronomer, while Newton's did not. OD. Kepler's approach was empirical, while Newton's was theoretical."

Answer:

1.19×10²² m/s

Explanation:

Kinetic Energy: This can be defined as the the energy of a body due to motion.

The formula for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass of proton. v = velocity of proton.

Making v the subject of the equation,

v = √(2Ek/m).................. Equation 2.

Given: Ek = 0.995×10⁻⁵ J, m = 1.673×10⁻²⁷ kg.

Substitute into equation 2

v = √(2×0.995×10⁻⁵/1.673×10⁻²⁷ )

v = 1.19×10²² m/s.

Therefore, the velocity of proton = 1.19×10²² m/s

Answer:985.71 Hz

Explanation:

Given

Speed of ambulance (source) [tex]v_s=108\ kmph\approx 30\ m/s[/tex]

Original Frequency of sound wave [tex]f=900\ Hz[/tex]

speed of sound waves [tex]v=345\ m/s[/tex]

According to Doppler apparent frequency will be different to original frequency when speed of source is moving w.r.t to observer.

[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]

where f'=apparent frequency

f=original frequency

[tex]v_o[/tex]=observer speed

[tex]f'=900\times (\frac{345+0}{345-30})[/tex]

[tex]f'=900\times (\frac{345}{315})[/tex]

[tex]f'=985.71\ Hz[/tex]        

Answer:

Explanation:

The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.

The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.

(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.

(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.

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Answer:

Explanation:

The thermal energy is given as

E=mc(T2-T1)

For the temperature change.

ΔT can be expressed in units of Kelvin or degrees Celsius. The ΔT of Kelvin is equal to that of ΔT of Celsius but not equal to the ΔT of Fahrenheit.

Therefore change in temperature of 100F is not equal to change in temperature of 100C

°F= 9/5 °C +32

So let assume 20°C, so the increase of 100°C will give 120°C.

Then °F = 68°F

Now the equivalent of 20°C is 68°F.

So let see the value of 120°C, which is the increase given.

°F= 9/5 ×120+32

°F= 248°F

The change in temp in Fahrenheit is 248-68=180°F

1°F change is = to 0.56°C change

Therefore, 100°F change in temperature in Fahrenheit is equal to 56°C in Celsius

Therefore the thermal energy is not the same, since they have are the same mass and specific heat but different changer in temperature.

So using the formulae given above, the thermal energy of the 100F change is greater than that of the 100C change..

A temperature drop of 100°C corresponds to a larger change in thermal energy than a drop of 100°F because 100°C equates to a more substantial temperature change (around 180°F). The container with the 100°C temperature drop, therefore, experiences a greater thermal energy change.

In comparing the change in thermal energy of liquids with a temperature drop of 100°F and 100°C, understanding that 1°C is equal to 1.8°F is crucial. To compare the energy change between the two, we must first recognize that the change of 100°F is approximately equivalent to a change of 55.5°C (100°F / 1.8 = 55.5°C).

Since the specific heat capacity is a property of a material that indicates how much heat energy it requires to change its temperature by a single degree, and this value is typically given in J/g°C, it is clear that a larger temperature change in Celsius would require more energy. Therefore, the container experiencing a drop of 100°C will require the removal of more thermal energy compared to the container with a drop of 100°F.

Answer:

a.-6.75 kgm/s

b.[tex]3375 N[/tex]

Explanation:

We are given that

Mass of baseball=0.15 kg

Initial speed=[tex]u=20m/s[/tex]

Final speed=[tex]v=-25m/s[/tex]

a.We know that

Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]

Momentum=[tex]mass\times velocity[/tex]

Using the formula

Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]

b.Time=[tex]2\times 10^{-3} s[/tex]

Force=[tex]\frac{Impulse}{time}[/tex]

Using the formula

Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N

Average force exerted by the bat on the ball=[tex]3375N[/tex]

Final answer:

The tension in the pendulum at the lowest point is 12.1 N. Without additional information, we cannot determine the exact angle at which the pendulum reaches its highest point. The tension in the pendulum at the highest point is 4.9 N.

Explanation:

Tension in the Pendulum Cable

To determine the tension in the pendulum cable at the lowest point, we can use the formula for circular motion.

T = mg + m(v^2/r)

Where:

g is the acceleration due to gravity (9.8 m/s²)

r is the length of the pendulum

Substituting the values given, we have:

T = 0.500 kg * 9.8 m/s² + 0.500 kg * (3.40 m/s)^2 / 0.80 m

Now, we can calculate the tension:

T = 4.9 N + 7.225 N

T ≈ 12.1 N

The Angle at the Highest Point

To find the angle at the highest point, we use the principle of conservation of energy. The kinetic energy at the lowest point equals potential energy at the highest point.

mgh = 1/2 mv^2

Solving for h gives us h = v^2/(2g), and then use trigonometry to find the angle θ.

h = r(1 - cosθ)

Without sufficient information to calculate the speed of the pendulum, we must leave the answer as an open calculation.

Tension at the Highest Point

At the highest point, the tension is equal to the weight of the pendulum alone since it is momentarily at rest, and there is no centripetal force required.

T = mg

Thus, the tension at the highest point will be:

T = 0.500 kg * 9.8 m/s² = 4.9 N.

Answer:

d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.

Explanation:

In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.

On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.

Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.

The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.

This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.

The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.

While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.

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Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

[tex]294.3sin\theta = 20[/tex]

[tex]sin\theta = 20/294.3 = 0.068[/tex]

[tex]\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0[/tex]

So the ramp cannot be larger than 3.9 degrees

Answer:

A & B are correct

Explanation:

Certain automotive sensors like crank shaft position sensor uses a.c. while most automotive components require DC to work properly through the battery which is incorporated with an alternator to keep it charged through a diode while continual supply of DC is sustained.

Technician A is incorrect as DC circuits are commonly used in automobiles, and Technician B is partially correct since some automotive sensors can generate an AC signal.

In the scenario presented, Technician A says that dc circuits are not normally used in automotive circuits, which is incorrect. DC circuits are indeed commonly used in automobiles for various functions such as starting the engine, powering lights, and running the dashboard instruments. Technician B says that certain automotive computer sensors use ac current. While most sensors in a car operate on DC current, there are some sensors, like variable reluctance sensors, that can generate an AC signal as they operate. Therefore, both statements are not entirely accurate; however, Technician B's statement has some validity regarding the existence of sensors that produce an AC signal within automotive applications.

Answer:

v = 31.5 m/s

Explanation:

given,

Speed of air balloon, u = 11.9 m/s

height of balloon above the ground, h = 43.4 m

acceleration due to gravity = 9.8 m/s²

Speed of the bag when it hit the ground = ?

using equation of motion

[tex]v^2 = u^2 + 2 g h[/tex]

[tex]v^2 = 11.9^2 + 2\times 9.8\times 43.4[/tex]

[tex]v= \sqrt{992.25}[/tex]

    v = 31.5 m/s

Speed of the bag before it hit the ground is equal to v = 31.5 m/s

Applying the physics of kinematics, specifically the equations of motion, the speed of the sandbag when it hits the ground can be calculated to be 29 m/s.

The speed of the sandbag when it hits the ground can be calculated with the concepts of kinematics in physics, in particular, the equations of motion. Considering that the positive direction is upwards and the balloon is moving upward at a velocity of 11.9 m/s, the initial velocity (vi) of the sandbag is +11.9 m/s. The sandbag is then acted upon by gravity, which imparts a downward acceleration of -9.8 m/s2. The final speed (vf), when the sandbag hits the ground, can be calculated using the equation vf2 = vi2 + 2ad, where a is acceleration and d is the distance. Substituting the given values: vf2 = (11.9 m/s)2 + 2*(-9.8 m/s2)*(-43.4 m), we get vf ≈ 29 m/s, which is the speed of the sandbag when it hits the ground.

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