The light-gathering power of a telescope is directly related to the area of the telescope's primary mirror. A mirror with four times the diameter of another mirror collects how many times more light as the smaller mirror does in the same amount of time?

Answer:

Swim in a strong north easterly direction so the the river carries him right across

Explanation:

The swimmer is on the west river bank with intentions of reaching the east river bank directly across him. If he just swims towards the east, the current of the river will carry him downstream landing him in a south easterly position, below where he intended to go. So the right thing to do would be to swim in a strong north easterly direction so the the river carries him right across.

To reach directly across the river from the west bank, the swimmer should head northeast to counteract the southward current of the river. This balancing act of velocities ensures they reach their desired location.

This question deals with the concept of Relative Velocity in Physics. When a swimmer attempts to cross a river, they must take into account the current of the river pushing them downstream. The velocity of the swimmer relative to the water needs to be added to the velocity of the river to get the resultant velocity, which describes the direction the swimmer actually travels relative to the shore.

In this scenario, if the current of the river is flowing due south, the swimmer on the west bank should not aim directly for the point across the river. They should aim slightly upstream to the north to counteract the force of the current. Hence, the swimmer should head in a northeasterly direction, which when combined with the river current, will get them directly across to the eastern shore.

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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]

Answer:

The temperature of the water increases because the nuclear reactor heats it producing steam

Explanation:

The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.

Answer:

a) speed of belt = 0.8114m/s

b) time required = 0.02secs

c) distance traveled = 0.016m

Explanation:

The detailed and step by step calculation is as shown in the attached files.

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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Answer:

a) Fa= Tension in the cable= 23010N

b) 19110N

c) 18232.5N

d=41.28m ,final velocity=0

Explanation:

a) Newtons 2nd law is given by Fnet=EF=ma

Fa= m(a + g)

Fa= 1950 (2+9.8)

Fa= 1950×11.8= 23010N

b) Fnet=0

Therefore Fb= W= 1950×9.8= 19110N

c) Fc= m(g - a)

Fc= 1950(9.8 - 0.45)

Fc= 18232.5N

d) First distance

ya= vt + 0.5at^2 = 0 + (0.5)(2)(2.05)^2

ya= 4.203m

yb= vt= 4.203×8=33.62m

yc = vt - (0.5at^2)

yc= 4.203×2.6) - (0.5×0.45×8^2)

yc = 10.93-14.4

yc =-3.46m

Dtotal = -3.46+33.62+4.203

Dtotal=41.28m

Explanation:

Below is an attachment of the solution.

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.

Car A will travel 150.1 km further before Car B overtakes it, and 30 seconds after Car B overtakes Car A, Car B will be 1.005 km ahead of Car A.

This is a physics problem involving relative speed and time. We are trying to find out how much farther Car A will travel before Car B catches up, as well as Car B's lead distance 30 seconds after it overtakes Car A.

a) How much farther will car A travel before car B overtakes it?

First, we calculate the Relative Speed of the two cars: Relative Speed = Speed of B - Speed of A = 121 km/h - 95 km/h = 26 km/h. Then we find out how long it takes for Car B to close that 41 km gap at this relative speed: Time = Distance/Speed = 41km / 26km/h = 1.58 hours. Therefore, in this duration, Car A will travel Distance = Speed * Time = 95km/h * 1.58h = 150.1 km.

b) How much ahead of A will B be 30 s after it overtakes A?

Here we just calculate the distance Car B travels in 30 seconds (converted to hours). Distance = Speed * Time = 121km/h * (30s / 3600s/h) = 1.005 km.

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Car A will travel approximately 150.1 km before Car B overtakes it. Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking. The solution involves calculating the time taken for Car B to close the initial gap and then the additional distance traveled in the subsequent 30 seconds.

Part (a)

To determine how much farther Car A will travel before Car B overtakes it, we can set up an equation based on the relative speeds of the two cars and the distance between them.

Let the time taken for Car B to overtake Car A be t hours.

In time t, Car A travels 95t km.

In time t, Car B travels 121t km.

At the point of overtaking, the distance covered by Car A plus the initial 41 km (since Car B started 41 km behind) will be equal to the distance covered by Car B:

95t + 41 = 121t

Solving for t:

121t - 95t = 41

26t = 41

t = 41/26 ≈ 1.58 hours

The distance Car A travels in this time is:

95 km/h × 1.58 hours ≈ 150.1 km

Part (b)

To find out how far ahead Car B will be 30 seconds after overtaking Car A:

Convert 30 seconds to hours:

30 seconds = 30/3600 hours = 1/120 hours.

In 1/120 hours :

Car A travels: 95 km/h × 1/120 hours ≈ 0.79 km.

Car B travels: 121 km/h × 1/120 hours ≈ 1.01 km.

The difference in distance traveled by Car B and Car A:

1.01 km - 0.79 km ≈ 0.22 km.

So, Car B will be approximately 0.22 km ahead of Car A 30 seconds after overtaking it.

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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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

Answer:

Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).

Explanation:

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.

Explanation:

Mechanical energy (the sum of potential and kinetic energy) is constant:

ME = PE + KE

ME = PE + ½ mv²

PE = ME − ½ mv²

So the slope of the line is -½ of the mass.

In a gravitational potential energy vs velocity^2 graph, the slope symbolizes the mass of the object. The graph represents the conversion of energies - kinetic and gravitational throughout a motion course. Peaks correspond to max potential energy and min kinetic energy, and vice-versa.

The slope of the gravitational potential energy vs the square of the velocity graph represents the object's mass in specific physical circumstances. This is because the equation for kinetic energy (which this graph is showing an indirect relationship with) is (1/2)mv^2, where m is mass and v is velocity. The slope can be interpreted as mass due to the rearrangement of this formula into the form used for this graph (Potential energy = 1/2 * m * v^2), wherein the slope (m) symbolizes the mass of the object.

Moreover, the information encapsulated in the potential energy graph as a whole represents energy conversion from kinetic to gravitational and vice versa. At the peak of a slope/motion course, potential energy is at its maximum and kinetic energy at its minimum. The opposite occurs at the bottom of the slope: both kinetic energy and potential energy reach their minimum values.

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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:

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.

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:

Solar wind particles can be captured by the Earth's magnetosphere. When these particles spiral down along the magnetic field into the atmosphere, they are responsible for?

Answer is Aurorae.

Explanation:

Solar Wind:

A planet's magnetic filed forms a shield to protect the surface of planet from energetic and charged particles coming from sun and other planets. The sun is continuously  sending out charged particles , called Solar wind.

Magnetosphere of planet:

If a planet has a magnetic field then it will interact with the solar wind and deflect the charged particles in the wind. Due to which an elongated cavity in solar wind formed. This cavity is called magnetosphere of the planet.

When Solar wind particles run in a magnetic field , they are  deflected and spiral down along magnetic field lines into the atmosphere.

Most of the solar wind particles deflected around the planet but a few particles manage to leak into the magnetic filed and become trapped in the magnetic field of the planet, to create Radiation Belts or Charged Particles Belts.

Variable Solar Wind can give some radiation belt particles with enough energy to spiral down along the magnetic filed into the atmosphere and Create Aurorae

Here Aurorae is an atmospheric phenomenon consisting of Bands, streamers of light, usually yellow , green or red , that move across the sky in the polar regions.

The charged particles from the solar wind spiral down along the Earth's magnetic field lines toward the poles, causing the atmospheric gases they collide with to emit light, resulting in auroras, commonly known as northern and southern lights.

When charged particles, known as the solar wind, emitted from the Sun encounter the Earth's magnetosphere, they can get trapped and follow the magnetic field lines towards the Earth's poles. Upon interacting with atmospheric gases, these charged particles cause a natural light display in the sky known as the auroras. These spectacular light shows are called the aurora borealis, or northern lights, in the Northern Hemisphere, and the aurora australis, or southern lights, in the Southern Hemisphere.

These auroras are more than just beautiful; they represent the interaction between the Sun's energy and our planet's protective magnetic field. During periods of intense solar activity, the solar wind is stronger, leading to more significant and vibrant auroral displays. These geomagnetic storms caused by solar activity can also have consequential effects on power grids, satellite operations, and communication systems.

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:

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:

Explanation:

Given

Voltage [tex]V=16\ V[/tex]

Work done [tex]W=1705\ J[/tex]

Work is Equivalent to energy

We know that Charge is given by

[tex]Q=I\cdot t[/tex]

where I=current

t=time

Energy [tex]E=P\times t[/tex]

P=power

P=V\times I

V=Voltage

I=current

Also Energy [tex]E=V\cdot I\cdot t[/tex]

[tex]I=\frac{E}{V\cdot t}[/tex]

Substitute the value of I in charge

[tex]Q=\frac{E}{V\cdot t}\times t[/tex]

[tex]Q=\frac{E}{V}[/tex]

[tex]Q=\frac{1705}{16}[/tex]

[tex]Q=106.56\ C[/tex]        

Answer:

Δβ = 28.2 dB

Explanation:

Attached is the explanation

Answer:

Potential difference will be equal to 2156.25 volt

Explanation:

We have given kinetic energy of electron [tex]KE=3.45\times 10^{-16}J[/tex]

Charge on electron [tex]e=1.6\times 10^{-19}C[/tex]

Let the potential difference is V

So energy electron will be equal to [tex]E=qV[/tex], here q is charge on electron and V is potential difference

This energy will be equal to kinetic energy of the electron

So [tex]1.6\times 10^{-19}\times V=3.45\times 10^{-16}[/tex]

V = 2156.25 volt

So potential difference will be equal to 2156.25 volt

Answer:

To me it would only make sense for it to be 51.33

Explanation:

w≈51.33

Length  3

Area  154

The width of the outside of the picture frame is 11 cm.

The distinction between a figure's length and width is that length denotes a figure's longer side, while width denotes its shorter side.

Given

L = w + 3

L w = 154

(w+3)w = 154

w² + 3 w - 154 = 0

(w+14)(w-11) = 0

w = 11 cm

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Answer:

The answer to your question is 56 ml

Explanation:

Data

V1 = 97 ml

P1 = 130 kPa

V2 = ?

P2 = 225 kPa

Formula

Use Boyle's law to solve this problem because it relates to the volume and the pressure.

                     V1P1 = V2P2

Solve for V2

                     V2 = (V1P1) / P2

Substitution

                     V2 = (97 x 130) / 225

Simplification

                     V2 = 12610 / 225

Result

                     V2 = 56 ml                      

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

Explanation:

Answer:

The last one

Explanation:

Answer:The Taller one

Explanation: Air drag(friction) also known as air resistance,the opposing force acting against the force of gravity, if a falling Object have a high air drag it will mean the object have a high air resistance. The higher the air resistance the higher the impact of the force of gravity on an object. The force of gravity will act to want to overcome the opposing force in order to bring the Object or the taller person down.

In a situation with no energy losses, an object sliding without rolling will reach the bottom of an incline first since all its energy goes into descending speed. For two mountain bikers with the same mass, the one with less air resistance will reach the bottom first.

The question concerns the concept of rolling motion versus sliding motion and its effect on the speed at which objects descend an incline. In an ideal scenario with no energy losses, an object that is rolling will have some of its energy in rotational kinetic energy in contrast to an object that is sliding which will only have translational kinetic energy. As a result, the object sliding without rolling, given the same initial conditions, will reach the bottom first due to having all of its energy contributing to its descent speed.

In the case of two mountain bikers with the same mass but differing in air resistance, the biker with less air drag will reach the bottom first. Air resistance acts as a form of friction that takes away some of the energy that could otherwise be converted into speed, thereby slowing down the biker.

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.

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.

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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."