Please help my grade depends on this

(a) The hanging cube moves a distance y to the floor. Determine whether the rotational kinetic energy of the stick-cube assembly is greater than, equal to, or less than the quantity 2mgy.

(b) Determine the angular speed of the stick-cube assembly.

(c) Determine the impulse applied to the stick-cube assembly.

(d) Suppose the swivel has significant mass. Determine whether the predicted answer in part (a) would be different. Explain your response.

Answer:

The type of collision, whether elastic or inelastic, can be determined by observing the behavior of the colliding objects before and after the collision. Here are some key characteristics that can help identify whether a collision is elastic or inelastic:

Conservation of Kinetic Energy: In an elastic collision, kinetic energy is conserved, while in an inelastic collision, some of the kinetic energy may be converted into other forms of energy.

Objects' Motion After Collision: In an elastic collision, objects bounce off each other and move independently, while in an inelastic collision, objects may stick together, deform, or move as a single mass.

Restitution Coefficient: In an elastic collision, the restitution coefficient is close to 1, indicating high bounce-back, while in an inelastic collision, the restitution coefficient is less than 1, indicating less bounce-back.

Conservation of Momentum: In both elastic and inelastic collisions, momentum is conserved, but the change in velocity of the objects after the collision can indicate whether the collision is elastic or inelastic.

If the period of the oscillator doubles, the wavelength of the waves on the string will also double and the wave speed does not change.

When an oscillator creates periodic waves on a stretched string and the period of the oscillator doubles, the following happens to the wavelength and wave speed:

1. Wavelength (λ): The relationship between the period (T), frequency (f), and wavelength (λ) is given by the equation:

T = 1/f

Since the period doubles (T becomes 2T), the frequency will halve (f becomes f/2) to maintain the relationship. The equation for the wave speed (v) is:

v = fλ

As the frequency is halved, to maintain the same wave speed, the wavelength must also double (λ becomes 2λ). So, the wavelength will increase.

2. Wave speed (v): In this situation, the wave speed remains constant. As mentioned above, when the frequency is halved, the wavelength doubles, which means the product of the frequency and wavelength (v = fλ) remains the same. Therefore, the wave speed does not change.

In summary, when the period of the oscillator doubles, the wavelength will double, and the wave speed will remain constant.

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The technique of interferometry allows two or more telescopes to obtain the angular resolution of a single telescope much larger than any of the individual telescopes.

This is achieved by combining the signals received by the telescopes to create a single image with a higher resolution. Interferometry is especially useful for studying objects with small angular sizes, such as stars and planets.

Additionally, interferometry allows astronomers to make astronomical observations without interference from light pollution, as it can separate the signals from the object being observed from the background light.

However, interferometry does not directly determine the chemical composition of stars, although it can provide information about their temperature and other physical properties.

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Assuming no air resistance, the maximum range that John's shotput can travel is achieved when it is launched at an angle of 45 degrees with respect to the ground.

This is known as optimal launch angle for a projectile thrown at a fixed speed. The actual distance traveled by the shotput will depend on its initial speed and other factors such as the height from which it is launched and local gravitational acceleration. The maximum range can be calculated using following formula:

R = (v^2/g) * sin(2θ)

where R is the maximum range, v is the initial speed of the shotput, g is the gravitational acceleration (assumed to be constant), and θ is the launch angle (measured from the horizontal).

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--The complete question is, Assuming no air resistance, how far will John's shotput travel and at what angle should he release it in order to achieve the maximum range, according to the principles of projectile motion?---

The negative sign indicates that the force is exerted in the opposite direction to the motion of the car. This force is applied over a short time interval and is relatively large, causing the car to experience a significant deceleration during the collision.

During the collision, the toy car experiences a change in momentum. Since momentum is conserved in the absence of external forces, the momentum of the car before the collision must be equal in magnitude and opposite in direction to the momentum after the collision.

The initial momentum of the car is given by:

p = mv = 0.3 kg * 0.82 m/s = 0.246 kgm/s

After the collision, the car comes to a stop, so its final momentum is zero. Therefore, the change in momentum is:

Δp = p_final - p_initial = -0.246 kg*m/s

The force exerted by the wall on the car during the collision can be calculated using the impulse-momentum theorem

J = Δp = FΔt

where J is the impulse, Δt is the time interval over which the force is applied, and F is the force

From the figure, we can see that the time interval for the collision is approximately 0.020 s. Therefore, the force exerted by the wall on the car is: F = Δp / Δt = -0.246 kg*m/s / 0.020 s = -12.3 N

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Based on my answer, I would expect collisions to be much more frequent at that time.

This is because factors such as higher traffic volume, increased speed, and lower visibility can contribute to a greater likelihood of collisions. Additionally, driver behavior, such as distraction or impatience, can also lead to more frequent collisions during peak times.

Collisions happen when two objects come into contact. In most cases, conservation of momentum and conservation of energy are used to solve collision-related problems. Any event where two or more bodies exert forces on one other quickly is referred to be a collision in physics.

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Since we don't know the mass of the satellite or its velocity, we can't solve for the mass of the planet with the given information alone. We would need at least one more piece of information to do so.

Answer -  Based on the given information, we can use the equation for gravitational force:

F = (G * m1 * m2) / r^2
where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Since we don't know the mass of the planet, we can't directly solve for it. However, we can use the fact that the satellite is in a circular orbit, which means that the gravitational force is equal to the centripetal force:
F = (m * v^2) / r
where m is the mass of the satellite and v is its velocity.
We can solve for m by rearranging the equation:
m = (F * r) / v^2

Now we can use this mass value and plug it into the original equation for gravitational force, along with the given values for r and F, to solve for the mass of the planet:
110 N = (G * m * m_planet) / (2.5x10^7 m)^2
m_planet = (110 N * (2.5x10^7 m)^2) / (G * m)

where G is a constant equal to 6.67x10^-11 N*m^2/kg^2.
Unfortunately, since we don't know the mass of the satellite or its velocity, we can't solve for the mass of the planet with the given information alone. We would need at least one more piece of information to do so.

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The weight of the object out of water is 800 kg.

To solve this problem, we need to use the principle of buoyancy. When an object is placed in water, it experiences an upward force called buoyant force, which is equal to the weight of the water displaced by the object.

In this case, the object has a volume of 1.00 m³, and 20.0% of its volume is above the waterline. Therefore, the volume of the object submerged in water is:

Vsubmerged = 1.00 m3 - 0.20 x 1.00 m³ = 0.80 m³

We also know the density of water is 1000 kg/m³. Therefore, the weight of the water displaced by the object is:

Wwater = density of water x volume of water displaced
Wwater = 1000 kg/m³ x 0.80 m³
Wwater = 800 kg

This means the buoyant force acting on the object is 800 kg. In order for the object to float, the buoyant force must be equal to the weight of the object. Therefore, we can find the weight of the object as:

Weight of object = Buoyant force = 800 kg

So the object weighs 800 kg out of the water.

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The period of the horizontal mass-on-a-spring system with a stiffness of 500 N/m and a mass of 25.7 kg is approximately 1.424 seconds.

We'll use the following terms in our calculation: stiffness of the spring (k), mass of the system (m), and period (T).

The formula to calculate the period of a mass-on-a-spring system is:

T = 2π √(m/k)

where:
T = period (in seconds)
m = mass of the system (25.7 kg)
k = stiffness of the spring (500 N/m)

Now, we'll plug in the values:

T = 2π √(25.7 kg / 500 N/m)

To calculate the square root:

T = 2π √(0.0514)

T = 2π × 0.2266

Finally, multiply by 2π:

T ≈ 1.424 seconds

So, the period of the horizontal mass-on-a-spring system is approximately 1.424 seconds.

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The frequency (in hertz) of the tuning fork is 485 Hz.

The speed of sound is given as 345 m/s and the wavelength is given as 71 cm. We need to find the frequency of the tuning fork.

We know that the speed of sound is equal to the product of frequency and wavelength:

Substituting the given values:

Solving for frequency:

frequency = 345 m/s ÷ 0.71 m

frequency = 485 Hz (rounded to the nearest whole number)

The frequency of a tuning fork is the number of vibrations or oscillations it makes per second and is typically measured in hertz (Hz). Tuning forks are commonly used in physics, music, and other fields to generate a pure tone with a specific frequency.

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The right response is resting tremor (option b). A patient may have spasticity, autonomic dysreflexia, and orthostatic hypotension following a spinal cord injury (SCI). SCI is not often linked to resting tremor.

SCI can interfere with the body's ability to communicate with the brain, leading to a variety of physical symptoms. Spasticity, which manifests as stiffness, muscle spasms, and increased muscle tone, is a frequent consequence. Patients with SCI at or above the T6 level may develop autonomic dysreflexia, a potentially fatal illness that is characterised by an abrupt rise in blood pressure. When someone stands up, their blood pressure drops, causing lightheadedness and dizziness. This condition is known as orthostatic hypotension.

While essential tremor, Parkinson's disease, and other neurological illnesses are frequently linked to resting tremor, SCI is not typically one of them.

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The right response is resting tremor (option b). A patient may have spasticity, autonomic dysreflexia, and orthostatic hypotension following a spinal cord injury (SCI). SCI is not often linked to resting tremor.

SCI can interfere with the body's ability to communicate with the brain, leading to a variety of physical symptoms. Spasticity, which manifests as stiffness, muscle spasms, and increased muscle tone, is a frequent consequence. Patients with SCI at or above the T6 level may develop autonomic dysreflexia, a potentially fatal illness that is characterised by an abrupt rise in blood pressure. When someone stands up, their blood pressure drops, causing lightheadedness and dizziness. This condition is known as orthostatic hypotension.

While essential tremor, Parkinson's disease, and other neurological illnesses are frequently linked to resting tremor, SCI is not typically one of them.

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If a solution absorbs green light, the colour observed will be its complementary colour, which is magenta, according to the transmission colour wheel.

Understanding the connection between the colour of light absorbed and transmitted is made easier with the help of the transmission colour wheel. This wheel predicts that if a substance absorbs one colour of light when it is transmitted, it will show the colour opposite.

As a result, magenta, which is green's complimentary colour, will be seen if a solution absorbs green light, which is in the centre of the colour wheel. This is because magenta, which is the transmitted colour, is situated on the transmission colour wheel exactly across from the green.

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The maximum distance the person can walk on the overhanging part of the plank before it just begins to tip is zero meters.

Let's consider the point where the plank just begins to tip. At this point, the weight of the plank and the person on it creates a clockwise moment, while the weight of the overhanging part of the plank creates an anticlockwise moment.

The weight of the plank is given as 225 N, and it acts at the center of the plank, which is 2.5 m from either end. Therefore, its clockwise moment is:

[tex]M1 = 225N*2.5m=562.5 Nm[/tex]

The weight of the person is 437 N, and it acts at a distance x from the right end of the plank. Therefore, its clockwise moment is:

M2 = 437 N * x

The weight of the overhanging part of the plank is:

[tex]W = 225N *(5.0m-1.1m)/2 = 618.75N[/tex]

It acts at a distance of (5.0 m - 1.1 m) / 2 = 1.95 m from the right end of the plank. Therefore, its anticlockwise moment is:

[tex]M3 = 618.75N*1.95m = 1207.81Nm[/tex]

At the point where the plank just begins to tip, these moments are balanced, so we have:

M1 = M2 + M3

[tex]562.5Nm = 437N* x+ 1207.81Nm[/tex]

Solving for x, we get:

[tex]x =(562.5 Nm - 1207.81 Nm)/437 N[/tex]

x = -1.51 m

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The intensity of a linearly polarized electromagnetic wave is directly proportional to the square of its electric field amplitude.

The electric field of a linearly polarized electromagnetic wave can be represented by a sine or cosine function, where the amplitude of the wave represents the maximum value of the electric field.

The intensity of the wave is proportional to the average power per unit area that is carried by the wave.

Mathematically, the intensity (I) of an electromagnetic wave is given by the formula:

I = (1/2)εcE0^2

where ε is the electric constant (approximately equal to 8.85 x 10^-12 F/m), c is the speed of light in a vacuum (approximately equal to 3.00 x 10^8 m/s), and E0 is the amplitude of the electric field.

From this formula, it is clear that the intensity of the wave is proportional to the square of the electric field amplitude.

Therefore, if the electric field amplitude of a linearly polarized electromagnetic wave is increased by a factor of 2, the intensity of the wave will increase by a factor of 4 (i.e., 2 squared).

Similarly, if the electric field amplitude is decreased by a factor of 2, the intensity of the wave will decrease by a factor of 4.

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Answer: hope it helps

Explanation:

A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.

The standing wave's fundamental frequency is the frequency of the first harmonic, which has one node and two antinodes, whereas the number of nodes determines the standing wave's harmonic number.

A 4.35 metre long, 137 gramme wire is being pulled at 125 newtons of force. With seven nodes total, including the endpoints, a standing wave has developed.

A collection of dots and dashes can be used to represent the wave pattern. The relationship between wave speed and wavelength is used to compute the standing wave's frequency. The tension in the wire and its linear mass density are used to calculate the wave speed.

The standing wave's fundamental frequency is the frequency of the first harmonic, which has one node and two antinodes, whereas the number of nodes determines the standing wave's harmonic number.

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Any electromagnetic radiation with a frequency lower than 9.84 x 10¹⁴ Hz (infrared, microwave, radio waves) will not cause the photoelectric effect in aluminum.

If infrared light shines on the aluminum surface, no electrons will be emitted via the photoelectric effect because the frequency of infrared light is lower than the threshold frequency of aluminum. The photoelectric effect occurs when a photon with enough energy (frequency) is absorbed by an electron in a metal, causing the electron to be emitted from the metal.

The minimum frequency or threshold frequency ([tex]f_{t}[/tex]) of a metal can be calculated using the equation:

[tex]f_{t}[/tex] = Φ ÷ h

where Φ is the work function of the metal (the minimum energy required to remove an electron from the metal) and h is Planck's constant. For aluminum, Φ = 4.08 eV.

Converting Φ to joules and using h = 6.626 x 10⁻³⁴ J s, we get:

Φ = 4.08 eV x 1.6 x 10⁻¹⁹ J/eV

Φ = 6.528 x 10⁻¹⁹ J

[tex]f_{t}[/tex] = 6.528 x 10⁻¹⁹ J ÷ 6.626 x 10⁻³⁴ J s

≈ 9.84 x 10¹⁴ Hz

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The visual rays reaching from the solar chromosphere are vanquished by rays at the red hydrogen Balmer Hα emission line wavelength. Thus, option A is correct.

The solar chromosphere is a thin coating of gas just beyond the photosphere, and it radiates most of its rays in the perceptible spectrum. The red hydrogen Balmer Hα emission line at a wavelength of 656.28 nm is notably vital in the chromosphere, and it overlooks the visual sunlight reaching from it.

This emission line is created by the growth of an electron in a hydrogen atom from the n-is 3 energy level to the n-is 2 energy grade, which radiates a photon with a wavelength of 656.28 nm.

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A 250-g ball approaching at 8.0 m/s suddenly collides head-on with a 550-g ball travelling at that same speed. The speed of the 250-g ball just after the collision is 5.5 m/s.

To start, we can figure out the system's overall momentum before the collision:

P_initial = m1v1 + m2v2

P_initial = (0.55 kg)(8.0 m/s) + (0.25 kg)(-8.0 m/s) [since the second ball is moving towards the first ball, its velocity is negative]

P_initial = 1.6 kg m/s

Next, we can use the conservation of kinetic energy to find the speed of the 250-g ball just after the collision:

[tex]1/2m_1v_1^2 + 1/2m_2v_2^2 = 1/2m_1v_1f^2 + 1/2m_2v_2f^2[/tex]

where v1f and v2f are the velocities of the two balls after the collision.

We can simplify this equation since the initial kinetic energy of the system is equal to the final kinetic energy of the system (since the collision is elastic):

[tex]1/2m_1v_1^2 + 1/2m_2v_2^2 = 1/2m_1v_1f^2 + 1/2m_2v_2f^2[/tex]

We can solve for v2f:

[tex]v_2f = \sqrt{((m_1v_1^2 + m_2v_2^2 - m_1*v_1f^2)/(m2))[/tex]

where we can use the equation for conservation of momentum to find v1f:

[tex]m_1v_1 + m_2v_2 = m_1v_1f + m_2v_2f\\\\v_1f = (m_1v_1 + m_2v_2 - m_2*v_2f)/(m_1)[/tex]

Plugging these values into the equation for v2f:

[tex]v2f = \sqrt{((m_1v_1^2 + m_2v_2^2 - m_1*((m_1v_1 + m_2v_2 - m_2*v_2f)/(m_1))^2)/(m_2))[/tex]

Simplifying this expression:

[tex]v_2f = \sqrt{((2m_1m_2*(v_1 - v_2)v_2f)/(m_2(m_1 + m_2)))[/tex]

v2f = sqrt((20.55 kg0.25 kg*(8.0 m/s - (-8.0 m/s))v2f)/(0.25 kg(0.55 kg + 0.25 kg)))

[tex]v_2f = 5.5 m/s[/tex]

Speed is characterised as the pace of motion or the amount of ground covered in a given amount of time. the speed of an object can change over time, which is referred to as acceleration. it has neither. Metres per second (m/s) or kilometres per hour (km/h) are the SI units for speed.

Since it enables us to characterise and examine an object's motion, speed is a fundamental notion in physics. The formula speed = distance/time can be used to determine an object's speed. For instance, an automobile would be moving at 50 km/h if it covered 100 kilometres in two hours. In addition to describing direction and velocity, speed may also be used to describe an object. The velocity vector's magnitude in this instance represents the speed.

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The nonconservative work done on the surfer is 2845.5 J.

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. In this case, we can calculate the initial and final kinetic energies of the surfer and find the difference, which will give us the net work done.

The initial kinetic energy of the surfer is:

[tex]K_i = (1/2) * m * v_i^2[/tex]

[tex]K_i = (1/2) * 73.2 kg * (1.44 m/s)^2[/tex]

K_i = 75.7 J

The final kinetic energy of the surfer is:

[tex]K_f = (1/2) * m * v_f^2[/tex]

[tex]K_f = (1/2) * 73.2 kg * (8.89 m/s)^2[/tex]

K_f = 2921.2 J

The change in kinetic energy is:

ΔK = K_f - K_i

ΔK = 2921.2 J - 75.7 J

ΔK = 2845.5 J

According to the work-energy theorem, this change in kinetic energy must be equal to the net work done on the surfer. Therefore, the nonconservative work done on the surfer is:

W_nc = ΔK

W_nc = 2845.5 J

So, the nonconservative work done on the surfer is 2845.5 J.

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The quantity of motion that occurs along a fault is termed C. displacement.

The quantity of motion that occurs along a fault is termed displacement. This refers to the distance and direction of movement that has taken place between two sides of a fault. It is measured in units such as meters or feet and is a crucial parameter in understanding the potential seismic hazard associated with a fault. Option A, the fault gouge, refers to the crushed and ground-up rock that accumulates along a fault zone due to movement. Option B, the fault gauge, is not a term commonly used in geology or seismology. Option D, accumulation, can refer to the build-up of stress or strain along a fault over time, which can eventually lead to displacement and an earthquake.

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Displacement is the amount of motion that happens along a fault. Hence option C is correct.

The displacement is the measure of the movement of the fault with respect to a particular point. It gauges the motion experienced during an earthquake or other types of fault activity.

The motion that happens during an earthquake or other fault activity is measured as displacement. Scientists can acquire insights into the physics of fault movement and increase their understanding of earthquake threats by examining displacement patterns during and after the earthquakes.

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The linear acceleration experienced by the bug depends on the time t. We cannot find a single value for it without knowing how much time has passed since the person started to pull the hose.

Since the bug is on the rim of the reel, it moves in a circular path along with the hose. Therefore, it experiences a centripetal acceleration that is given by the formula:

[tex]a = r * ω^2[/tex]

where r is the radius of the circular path, and ω is the angular velocity of the bug.

Initially, when the reel is at rest, the angular velocity of the bug is zero. When the person starts to pull the hose with a linear acceleration of [tex]0.75 m/s^2,[/tex] the reel also starts to rotate with an angular acceleration of:

α = a / r = [tex](0.75 m/s^2)[/tex] / (0.10 m) = [tex]7.5 rad/s^2[/tex]

Using the formula for angular acceleration, we can find the angular velocity of the reel after a certain time t:

ω = α * t

The angular velocity of the bug is the same as that of the reel, so we can use the same formula to find the angular velocity of the bug after time t.

Once we know the angular velocity of the bug, we can use the formula for centripetal acceleration to find the linear acceleration experienced by the bug:

[tex]a = r * ω^2[/tex]

Substituting the given values, we get:

a = [tex](0.33 m) * (α * t)^2[/tex]

a = [tex](0.33 m) * [(7.5 rad/s^2) * t]^2[/tex]

a = [tex]18.56 t^2 m/s^2[/tex]

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The intensity of the radiation at the location of the sensor is approximately 6.58 x 10^3 W/m².

To calculate the intensity of the radiation at the given location, we need to consider the relationship between the electric field and intensity. The terms involved in this question are distance, electric field, and intensity.

Intensity (I) is the power (P) of the radiation distributed over a specific area (A). In this case, we are given the root-mean-square (RMS) value of the electric field (E) at a distance (d) of 29.7 cm from the light bulb.

The relationship between electric field and intensity can be described using the formula:

I = (E^2) / (2 * ε₀)

Where ε₀ is the vacuum permittivity, which has a value of approximately 8.85 x 10^-12 F/m.

Now, let's plug in the given values:

I = (341.4 V/m)^2 / (2 * 8.85 x 10^-12 F/m)

I = (116556.36 V²/m²) / (1.77 x 10^-11 F/m)

I = 6.58 x 10^3 W/m²

Therefore, the intensity of the radiation at the location of the sensor is approximately 6.58 x 10^3 W/m². This value represents the power of the radiation emitted by the light bulb distributed over a square meter at a distance of 29.7 cm from the bulb.

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Answer:

The answer is Continental Grip

The peak electric field 10 km from the transmitter is approximately 15.435 mV/m.

To find the peak electric field 10 km from the transmitter, we can use the inverse square law.

This law states that the intensity of a wave (such as the electric field in this case) is inversely proportional to the square of the distance from the source.

Here's a step-by-step explanation:

1. Note the initial distance (d1) and electric field (E1):

d1 = 2.1 km, E1 = 350 mV/m.

2. Convert d1 to meters:

d1 = 2100 m.

3. Note the final distance (d2):

d2 = 10 km.

4. Convert d2 to meters:

d2 = 10,000 m.

5. Use the inverse square law formula:

E2 = E1 * (d1²) / (d2²).

6. Plug in the values:

E2 = 350 * (2100²) / (10,000²).

7. Calculate E2:

E2 ≈ 15.435 mV/m.

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The reasons that a promontory will be more vulnerable to wave erosion than a bay;- Waves bend around a promontory and strike it from both sides,- Powerful waves focus most of their energy at a promontory and - A promontory will receive more wave action than a bay.

A promontory is more vulnerable to wave erosion than a bay due to the following reasons:

1. Waves bend around a promontory and strike it from both sides: This phenomenon, called wave refraction, concentrates the wave energy on the promontory, making it more prone to erosion.

2. A promontory will receive more wave action than a bay: Bays are generally more sheltered and have a lower exposure to waves, whereas promontories are exposed to the full force of waves, leading to more erosion.

3. Powerful waves focus most of their energy at a promontory: Due to the shape of the coastline, waves tend to focus their energy on the headlands, like promontories, which makes them more vulnerable to erosion compared to bays.

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The option A is  best representation of the path that the ball would take after the string breaks.

When the string of a pendulum breaks, the ball's path will follow the laws of motion, specifically the law of conservation of energy. As the ball was halfway to its highest position, it had a certain amount of potential energy.

When the string broke, this potential energy would convert to kinetic energy, causing the ball to move in a straight line tangent to the point where the string broke.

Therefore, the path that the ball would take after the string breaks would be a straight line away from the pivot point of the pendulum, as shown in option A. The other paths shown do not follow the laws of motion and do not account for the conservation of energy. Option (A) is the correct answer.

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Note the full question is

A pendulum is swinging upward and is halfway toward its highest position, as shown, when the string breaks. which of the paths shown best represents the one that the ball would take after the string breaks?

A) A

B) B

C) C

D) D

E) E

The tension in the rope pulling the skier is 514 N.

The power produced by the motor to maintain a constant speed of 13.4 m/s is 35700 W. The power required to pull the skier at the same constant speed is 36800 W. The difference in power is due to the additional work required to overcome the frictional force between the skier and the water.

The force required to maintain a constant speed can be calculated using the formula F = P/V, where F is the force, P is the power, and V is the velocity. For the motor to maintain a constant speed of 13.4 m/s, the force required is 2672.69 N.

To find the tension in the rope pulling the skier, we need to subtract the force required to maintain the constant speed (2672.69 N) from the force required to pull the skier (which we do not know yet). The result is the tension in the rope pulling the skier.

To find the force required to pull the skier, we can use the same formula, F = P/V, with the power of 36800 W and the velocity of 13.4 m/s. This gives a force of 2746.27 N.

Subtracting the force required to maintain the constant speed (2672.69 N) from the force required to pull the skier (2746.27 N) gives a tension in the rope of 514 N.

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