Which statements are true about the amount of gravitational potential energy an object has?
The amount increases as the object is lifted higher.
The amount varies according to the material of the object.
The amount varies according to the mass of the object.
The amount increases the more quickly the object is lifted.

Answers

Answer 1

Answer:

The statement "The amount increases as the object is lifted higher" is true about the amount of gravitational potential energy an object has. The statement "The amount varies according to the material of the object" is false. The amount of gravitational potential energy only depends on the mass of the object and its elevation from the reference point. The statement "The amount varies according to the mass of the object" is true. The more massive an object is, the more gravitational potential energy it has. The statement "The amount increases the more quickly the object is lifted" is false. The amount of gravitational potential energy only depends on the object's height above the reference point, not the speed at which it is lifted.

Answer 2

Answer: (A)

Explanation: The amount increases as the object is lifted higher.


Related Questions

When a projectile is launched at an angle  from a height h1 and the projectile lands at the same height, the maximum range, in the absence of air resistance, occurs when  = 45. The same projectile is then launched at an angle  from a height h1, but it lands at a height h2 that is higher than h1, but less than the maximum height reached by the projectile when  = 45. In this case, in the absence of air resistance, does the maximum range still occur for  = 45? All angles are measured with respect to the horizontal direction.

Answers

No, the maximum range will not occur for θ = 45°. The maximum range occurs when the horizontal component of the initial velocity is equal to the initial vertical velocity at the instant the projectile hits the ground.

When a projectile is launched at an angle θ from a height h1, the time it takes to reach the highest point of its trajectory is given by t = (V0 sin θ)/g, where V0 is the initial velocity and g is the acceleration due to gravity. At the highest point, the vertical component of the velocity is zero and the horizontal component of the velocity remains constant. The time of flight of the projectile is given by T = (2V0 sin θ)/g.

The range of the projectile is given by R = (V0 sin 2θ)/g, where sin 2θ = 2sin θ cos θ. When the projectile lands at a height h2 that is higher than h1, the maximum range will occur at an angle that depends on the initial velocity and the difference in height between the launch and landing points. In general, the maximum range will occur at an angle that is less than 45° when the landing height is greater than the launch height, and at an angle that is greater than 45° when the landing height is less than the launch height. The specific angle can be calculated using the equation for range and setting the derivative with respect to θ equal to zero.

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Can someone please help me and thank you

Answers

Answer:

D.reflected waves of the same wavelength and less intensity than the original sound source

Being over stressed can make you feel

1. Excited

2. Happy

3. Irritable

4. Relaxed

Answers

Answer: The answer would be 4.

Explanation:

Being stressed is not a good feeling. Since None of the other answers suit the description of "overstressed", it would be 4.

irritable

Explanation:

irrtable meaning Easily annoyed or easily made mad

A ball thrown with an initial velocity of u(10i+15j)m/s when it reaches the top of it trajectly neglucating air resistance what is avelocity & acceleration ?

Answers

At the apex of the trajectory, the ball's velocity and acceleration are u (10i+15j) m/s and zero

The initial velocity of the ball is u (10i+15j) m/s.

Velocity at the top of the trajectory: Since the ball is thrown with an initial velocity, it will reach the top of its trajectory with the same velocity, since there is no air resistance. Therefore, the velocity of the ball at the top of the trajectory is u (10i+15j) m/s.

Acceleration: The acceleration of the ball at the top of the trajectory is zero, since the ball is not accelerating (there is no acceleration due to air resistance).

Therefore, the velocity and acceleration of the ball at the top of the trajectory are u (10i+15j) m/s and zero, respectively.

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How is the momentum conversation connected to Newton's Second Law? Why is an external force important here?

How are the conservation of momentum and Newton's Third Law connected? How can they really say the same thing?

Answers

The conservation of momentum is a fundamental principle in physics that is closely related to Newton's second law of motion. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass:

F = ma

This law can be rearranged to give:

a = F/m

If we apply a force to an object, it will accelerate in the direction of the force. However, if there is no external force acting on a system of objects, the total momentum of the system is conserved.

The conservation of momentum is connected to Newton's second law through the concept of external forces. According to Newton's second law, an external force is required to change the momentum of an object. If there are no external forces acting on a system of objects, the total momentum of the system cannot change.

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Urgent help! Two pure tones Cs and Gs, with frequencies from the Pythagorean diatonic scale, are sounded simultaneously. Find a) the frequencies of the three combination tones and b) the notes on the Pythagorean scale to which these tones belong.

Answers

[tex]answer[/tex]

When two pure tones with frequencies f1 and f2 are sounded simultaneously, combination tones are produced at frequencies of (f1 + f2) and (|f1 - f2|). In this case, the two pure tones are Cs and Gs from the Pythagorean diatonic scale, which have frequencies of 256 Hz and 384 Hz, respectively.

a) The frequencies of the three combination tones are:

- (f1 + f2) = 640 Hz (Cs + Gs)

- (|f1 - f2|) = 128 Hz (Gs - Cs)

- (2f1 - f2) = 128 Hz (Cs + Gs - 2Cs)

b) To find the notes on the Pythagorean scale to which these tones belong, we need to compare their frequencies to those of the Pythagorean diatonic scale. The Pythagorean diatonic scale is based on a ratio of 3:2 between adjacent notes, which means that the frequency of each note is 3/2 times the frequency of the previous note. Starting from Cs (256 Hz), the frequencies of the Pythagorean diatonic scale are:

- Ds: 288 Hz (256 Hz x 3/2)

- Es: 324 Hz (288 Hz x 3/2)

- Fs: 341.33 Hz (324 Hz x 81/80)

- Gs: 384 Hz (341.33 Hz x 3/2)

- As: 432 Hz (384 Hz x 3/2)

- Bs: 486 Hz (432 Hz x 3/2)

- Cs: 512 Hz (486 Hz x 81/80)

Comparing the combination tones to the Pythagorean diatonic scale, we can see that:

- 640 Hz is between Gs (384 Hz) and As (432 Hz)

- 128 Hz is between Cs (256 Hz) and Ds (288 Hz)

- 128 Hz is between Cs (256 Hz) and Ds (288 Hz)

Therefore, the combination tones belong to the notes Gs-As, Cs-Ds, and Cs-Ds, respectively.

Two physics students are doing a side competition during a game of bowling, seeing who can toss a ball with the larger momentum. The first bowler throws a 4.5 kg
ball at 8.6 m/s

A second bowler throws a 6.4 kg ball. What speed must she beat to win the competition?

Answers

The speed that must be beat to win the competition is 6.05 m/s.

What is Speed?

Speed is the rate of change of distance.
To calculate the speed the second bowler must beat to win the competition, we use the formula below

Formula:

v = mV/M.................. Equation 1

Where:

v = Speed the second bowler must beatm = Mass of the first bowlerM = Mass of the second bowlerV = Speed of the first ball

From the question,

Given:

m = 4.5 kgV = 8.6 m/sM = 6.4 kg

Substitute these values into equation 1

v = (4.5×8.6)/6.4v = 6.05 m/s

Hence, the speed that must be beat is 6.05 m/s.

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your hand feels cold when you hold an ice cube because A. the ice radiates cold to your hand B. the ice conducts cold to your hand C. your hand cools down by convection. D.Your hand transfers thermal energy to the ice​

Answers

your hand feels cold when you hold an ice cube because D.Your hand transfers thermal energy to the ice

Is there a flow of cold from the ice to your hand?

The ice does not transfer cold to your hand. Heat is transferred from your hand to the ice. When you are transmitting heat to the metal, it feels chilly to the touch.

Heat energy constantly flows from high temperature to low temperature. The temperature of the fingers is higher than that of ice. As a result, energy will be transferred from the finger to the ice.

This is due to the larger quantity of heat from our body moving to the ice through our skin (through thermal conduction) in order to achieve thermal equilibrium between the ice and the body.

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what is The possibility of telepathy without MRI machines?

Answers

Some people believe in the possibility of telepathy through paranormal or supernatural means, but these claims are not supported by empirical evidence.

What is MRI?

MRI stands for Magnetic Resonance Imaging. It is a medical imaging technique that uses strong magnetic fields and radio waves to generate detailed images of internal structures of the body. MRI machines produce high-quality images of the body's soft tissues and organs, making them useful in diagnosing and monitoring various medical conditions. MRI scans are often used to detect abnormalities in the brain, spine, joints, and other parts of the body.

Telepathy, the ability to communicate thoughts or ideas through means other than the known senses, is not currently supported by scientific evidence. While MRI machines and other advanced technologies are used to study brain activity and communication, the concept of telepathy without such devices is not currently recognized as a scientifically valid phenomenon.

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(a)
Calculate the force (in N) needed to bring a 900 kg car to rest from a speed of 85.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).


(b)
Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

Answers

(a) To calculate the force needed to bring the car to rest, we can use the equation:

f = (m * v^2) / (2 * d)

where:
m = 900 kg (mass of the car)
v = 85.0 km/h = 23.6 m/s (initial velocity of the car)
d = 110 m (stopping distance)

Plugging in the values, we get:

f = (900 kg * (23.6 m/s)^2) / (2 * 110 m)
f = 23095.91 N or 2.31 x 10^4 N

Therefore, the force needed to bring the car to rest from a speed of 85.0 km/h in a distance of 110 m is approximately 23,095.91 N or 23.1 kN.

(b) If the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m, we can calculate the force exerted on the car using the formula:

f = m * a

where:
m = 900 kg (mass of the car)
a = -v^2 / (2 * d) = -(23.6 m/s)^2 / (2 * 2.00 m) = -2793.2 m/s^2 (acceleration of the car)

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity of the car.

Plugging in the values, we get:

f = 900 kg * (-2793.2 m/s^2)
f = -2.51 x 10^6 N or -2.51 MN

Therefore, the force exerted on the car when it hits a concrete abutment at full speed and is brought to a stop in 2.00 m is approximately 2.51 MN (or 2510 kN), which is much greater than the force found in part (a). This is because the stopping distance is much shorter, so the deceleration (and therefore the force) must be much greater to bring the car to a stop in the same amount of time.

One end of a spring with a spring constant of 109 N/m is held firmly in place, and the other end is attached to a block with a mass of 2.13kg. The block undergoes SHO (simple harmonic motion) with no friction. At time t = 0.6763s, the position and velocity of the block are:

x(0.6763s) = -0.1031m

y(0.6763s) = 0.5303m/s


A. What was the position, in meters, at t = 0.00s?

B. What was the velocity, in meters per second, at t = 0.00s?

Answers

At time zero, the location is -0.1031m, and the speed is 0 m/s.

How can you determine a block's top speed?

How fast can the block go at its most. When the spring reaches its equilibrium length, all of the stored energy in the spring is transferred to kinetic energy, and the block accelerates to its maximum speed. Since the item is at rest, Ki = 0. As there is no friction, Wnc = 0.

A = x(0.6763s) = -0.1031m

v(0.6763s) = -Aω sin(ω*0.6763s) = 0.5303 m/s

ω = sqrt(k/m) = sqrt(109 N/m / 2.13 kg) = 5.148 rad/s

Now we can use these values to find the position and velocity at t = 0:

A cos(ωt) = A cos(0) = A = -0.1031m

-v(0) = Aω sin(ωt) = -Aω sin(0) = 0

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Match the model of the atom to its description.

Question 3 options:

Now considered inaccurate because it does not include the nucleus


Most accurate and newest model, but often too complicated for use in the classroom


Electrons are orbiting the nucleus in this model. This model is useful and used often, but not fully accurate

1.
"Plum pudding" or "raisin bread" model of the atom

2.
Bohr's planetary model of the atom

3.
The quantum or cloud model of the atom

Answers

"Plum pudding" or "raisin bread" model of the atom

Bohr's planetary model of the atom

The quantum or cloud model of the atom

What is Plum Pudding Model?

The "Plum pudding" or "raisin bread" model of the atom is a model where electrons are distributed throughout the atom and the positive charge is spread out to create a sort of "pudding" with "raisins" of negative charge. This model was proposed by J.J. Thomson in the early 1900s and is now considered inaccurate because it does not include the nucleus.

Bohr's planetary model of the atom is a model where electrons orbit the nucleus in specific energy levels or "shells". This model was proposed by Niels Bohr in 1913 and was an improvement over the "Plum pudding" model. It is useful and used often in introductory chemistry, but it has limitations in accurately describing more complex atoms.

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You are measuring the current in a circuit that is operated on an 18 V battery. The
ammeter reads 40 mA. Later you notice the current has dropped to 20 mA. How much has the voltage changed

Answers

The voltage has not changed from the initial reading of 18 V. The change in current is due to the change in the resistance of the circuit, and not a change in voltage.

What is Voltage?

Voltage, also known as electric potential difference, is a measure of the electrical potential energy per unit charge in an electrical circuit. It is often represented by the symbol "V" and is measured in volts (V).

Assuming the circuit is a constant resistance, Ohm's law states that the current in a circuit is directly proportional to the voltage. Therefore, we can use Ohm's law to calculate the resistance of the circuit:

Resistance = Voltage / Current

Using the initial readings:

Resistance = 18 V / 40 mA = 450 ohms

Using the later readings:

Resistance = 18 V / 20 mA = 900 ohms

Therefore, the resistance of the circuit has increased from 450 ohms to 900 ohms.

We can calculate the change in voltage by using Ohm's law and the new resistance value:

Voltage = Current x Resistance

Using the later readings:

Voltage = 20 mA x 900 ohms = 18,000 mV or 18 V

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A 3.30-m-long, 560 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kg construction worker stands at the far end of the beam.

What is the magnitude of the torque about the point where the beam is bolted into place?

Answers

the magnitude of the torque about the point where the beam is bolted into place is 2232.17 N⋅m.

The torque about the point where the beam is bolted into place can be calculated as:

τ = Fd sinθ

where F is the force, d is the distance from the point where the force is applied to the pivot point, and θ is the angle between the force vector and the position vector.

In this case, the force is the weight of the construction worker, which can be calculated as:

F = mg

where m is the mass of the worker and g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]).

Substituting the given values, we get:

F = (69.0 kg)(9.81 m/[tex]s^2[/tex]) = 676.89 N

The distance d can be calculated as the length of the beam from the pivot point to the point where the worker is standing:

d = 3.30 m

The angle θ between the force vector and the position vector is 90 degrees, since the force is acting perpendicular to the beam.

Substituting the values in the torque equation, we get:

τ = Fd sinθ = (676.89 N)(3.30 m)(sin 90°) = 2232.17 N.

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Need help on this question!!!

True or False. Animals can be at the top of more than one food chain.

Answers

Answer:

false

Explanation:

Can animals be at the top of more than one food chain? There cannot be too many links in a single food chain because the animals at the end of the chain would not get enough food (and hence, energy) to stay alive.

147 9001 of energy is added to a certain mass of ethyl alcohol liquid at 18°C. This is the amount of energy used to completely evaporate all the liquid. Use the information below to determine the minimum mass of the ethyl alcohol.​

Answers

The minimum mass of the ethyl alcohol liquid is 37.57 g.

First, we need to determine the heat of vaporization of ethyl alcohol at 18°C. The heat of vaporization of ethyl alcohol varies with temperature, so we need to find the value at 18°C. We can use the Clausius-Clapeyron equation:

ln(P₂/P₁)= -ΔHvap/R * (1/T₂ - 1/T₁)

where P₁ is the vapor pressure of ethyl alcohol at its boiling point (78.4°C), P₂ is the vapor pressure at 18°C, ΔHvap is the heat of vaporization, R is the gas constant (8.314 J/mol·K), T₁ is the boiling point temperature, and T₂ is the temperature at which we want to find the heat of vaporization.

Rearranging the equation to solve for ΔHvap:

ΔHvap = -R * (1/T₂ - 1/T₁) * ln(P₂/P1)

Substituting the known values:

ΔHvap = -8.314 J/mol·K * (1/291.15 K - 1/351.55 K) * ln(13300 Pa/76000 Pa) = 39,343 J/mol

Next, we need to use the heat of vaporization to calculate the minimum mass of the liquid. The given amount of energy (1479001 J) is the amount of energy required to completely evaporate all the liquid, so it is equal to the product of the mass of the liquid and its heat of vaporization:

1479001 J = m * 39343 J/mol

Solving for m:

m = 37.57 g

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A phone weighs a total of 5kg and throws it across the room at 5 m/s because he's upset it isn't spring break yet. What is the kinetic energy
of the phone

Answers

The phone has a kinetic energy of 62.5 joules.

How do you calculate kinetic energy?

The formula for kinetic energy is provided as K E = 1 2 m v 2. Where m is the body's mass, v is its motion, and KE is its kinetic energy.

You can determine the phone's kinetic energy (KE) by using the following formula:

KE = 1/2 * m * v²

where m is the mass of the phone and v is its velocity.

In this case, the mass of the phone is 5 kg and its velocity is 5 m/s. With these numbers entered into the formula, we obtain:

KE = 1/2 * 5 kg * (5 m/s)²

KE = 1/2 * 5 kg * 25 m²/s²

KE = 62.5 J

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Why is it dangerous for an archer to walk around with an arrow nocked and bowstring pulled to tension

Answers

It is important for archers to follow proper safety protocols and always keep their bows and arrows pointed in a safe direction and not to nock an arrow until they are ready to shoot.

It is dangerous for an archer to walk around with an arrow nocked and bowstring pulled to tension because it creates a situation where the bow is essentially loaded and ready to fire. Any accidental bump or movement could cause the arrow to be released, potentially causing injury to the archer or anyone nearby. This can be especially dangerous if the bow is pointed towards people or valuable objects.

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How many waves are on this string? ​

Answers

A standing wave is composed of two waves with the same frequency and amplitude traveling in opposite directions and interfering with each other.  Thus we have to waves here.

Is a standing wave composed of two waves?

These waves are known as the incident wave and the reflected wave. When the incident wave and the reflected wave interfere constructively, they create points of maximum displacement known as antinodes.

When they interfere destructively, they create points of minimum displacement known as nodes. The pattern of antinodes and nodes remains fixed in space, giving rise to a standing wave.

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In the most common form of colorblindness, a cone system malfunctions, rendering the color
indistinguishable from other color combinations.

Answers

The most common form of color blindness is called red-green color blindness.

This type of color blindness is caused by a malfunction of the red and green cone systems in the eye, which are responsible for detecting different wavelengths of light. This malfunction causes individuals to have difficulty distinguishing between certain combinations of red and blue, as they appear to be the same color or shade to the affected person. The exact cause of this malfunction is still unknown, but is believed to be caused by genetic factors, environmental influences, or a combination of both.This color blindness affects up to 8% of men and 0.5% of women, and is caused by a mutation in the gene that codes for the red-sensitive cone cells in the eye.

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complete question:in the most common form of color blindness, the ___ cone system malfunctions rendering ____ indistinguishable from certain combinations of blue and red

For a pith ball experiment set up on Earth, which image below correctly depicts the result if you gave the left-hand pith ball a charge of +7.0 × 10-^6C and the right-hand pith ball a charge of +11.0 × 10^-6 C? Assume the pith balls have equal mass.

Answers

The two pith balls with like charge will repel each other and move away from each other until the electric force between them is balanced by the tension force of the string holding them up. Hence, the correct option is (A).

The left-hand pith ball has a charge of +7.0 × 10⁻⁶C and the right-hand pith ball has a charge of +11.0 × 10⁻⁶ C, they will both repel each other due to their like charges. The magnitude of the force between them will depend on the distance between them and the amount of charge on each ball, according to Coulomb's law. The pith balls will move away from each other until the electric force between them is balanced by the tension force of the string holding them up. The resulting position of the pith balls will depend on the angle of the string relative to the vertical direction. The greater the difference in charge between the two balls, the greater the angle will be.

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Starting at t = 0, a horizontal net force F⃗ =(0.290N/s)ti^+(−0.445N/s2)t2j^
is applied to a box that has an initial momentum p⃗ =(−3.10kg⋅m/s)i^+(4.10kg⋅m/s))j^
.
a) What is the x-component of the momentum of the box at t = 2.00 s ?
b) What is the y-component of the momentum of the box at t= 2.00 s ?

Answers

a) The x-component of momentum is given by pₓ = m * vₓ, where m is the mass of the box and vₓ is the x-component of velocity. Since the force is only acting in the x-direction, there is no net force in the y-direction, and the y-component of momentum is conserved.

Therefore, we can use the initial y-component of momentum to find the y-component of momentum at any time.

At t = 2.00 s, the x-velocity can be found by integrating the x-component of the force:

Fₓ = (0.290 N/s)t

vₓ = ∫Fₓ dt = (0.145 N/s)t² + C

Using the initial condition that vₓ(0) = pₓ(0)/m = (-3.10 kg⋅m/s) / (m), we can solve for C:

C = -3.10 m/s

Therefore, vₓ = (0.145 N/s)t² - 3.10 m/s

And the x-component of momentum at t = 2.00 s is:

pₓ = m * vₓ = m * [(0.145 N/s) (2.00 s)² - 3.10 m/s]

pₓ = (550 kg) * [-1.54 m/s] ≈ -847 kg⋅m/s

b) As mentioned above, the y-component of momentum is conserved since there is no net force in the y-direction. Therefore, the y-component of momentum at t = 2.00 s is the same as the initial y-component of momentum, which is:

pᵧ = m * vᵧ = m * [(4.10 kg⋅m/s) / (m)] = 4.10 kg⋅m/s

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An object that is fired horizontally hits the ground in 0.500 seconds. If it had been fired with twice the speed, but still horizontally, it would have hit the ground in

Answer choices:

A) Less than 0.500s
B) More than 0.500s
C) 0.500s
D) Not enough Information

Answers

The answer is (C), 0.500s.

This is because in both instances the bullet is fired with all horizontal speed, meaning the initial vertical component of velocity ([tex]\vec v_{y}[/tex]) will equal 0. We also know that vertical acceleration ([tex]\vec a_{y}[/tex]) is the acceleration of gravity, which is 9.8 m/s^2. In both cases when the bullet is fired, the acceleration of gravity will pull them down in the same way. Thus, the time the bullet takes to hit the ground will be the same.

In short, because the velocity and acceleration is constant in the vertical direction, the bullets will fall in the same amount of time.

a sound wave takes 6 seconds to complete one cycle. what is the frequency of this wave

Answers

Therefore, the frequency of the sound wave is 0.1667 Hz.

How lengthy is a wave at 20 Hz?

Since we can fit 20 cycles into a distance of 340 meters, the wavelength for 20 Hz is equal to 340 metres divided by 20, or 17 metres. The wavelength is described as the length of this pattern for one cycle.

The quantity of cycles per second is used to describe a wave's frequency. The sound wave in this instance goes through one rotation in 6 seconds.

So the frequency of the wave can be calculated as:

Frequency = 1 / Time period

where Time period is the time taken to complete one cycle.

Therefore, the frequency of the sound wave can be calculated as:

Frequency = 1 / 6 seconds = 0.1667 Hz

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An ocean wave travels at approximately 1.97 m/s. This is 4 miles per hour. The frequency of the waves is roughly 0.07 hz (or roughly 4 waves per minute). What is the Wavelength?

Answers

Therefore, the wavelength of the ocean wave is approximately 28.14 meters.

What is the equation for a wave's frequency?

The equation f=v f = v, where  is the wavelength in metres and v is the wave speed in m/s, can be used to determine the frequency of a wave if the radius and speed of the wave are known. This also provides the wave's frequency in Hertz.

The wave speed calculation can be used to calculate the ocean wave's wavelength:

v = λf

where the wavelength is, the frequency is f, and v is the wave speed.

The wave frequency is given as 0.07 Hz and the wave speed is provided as 1.97 m/s. We can use the following conversion ratio to change 4 miles per hour to metres per second:

1 mile/hour = 0.44704 m/s

So, 4 miles/hour = 1.78816 m/s

Now we can substitute the values into the wave speed equation and solve for λ:

λ = v/f = 1.97 m/s / 0.07 Hz = 28.14 m

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Hi! I need urgent help with all parts of the question which is attached below.

Answers

The force that causes an object to move in circular motion is called the centripetal force. It is always directed towards the center of the circle

How does an object move in circular motion?

We have to convert from rev/s to rad/s and we have 7.5 rad/s

Using;

v = ωr

v = 7.5 rad/s * 0.05 m

= 0.375 m/s

The centripetal force = mv^2/r

= 0.02 * (0.375)^2/0.05

=0.056 N

As the object moves around the circle, it experiences a constant change in direction, but its speed can remain constant or change. When the speed remains constant, the centripetal force and the force of the object's motion balance each other, resulting in uniform circular motion.

The object may deform or change its shape as it rotates at a higher speed. For example, a rubber ball may stretch or flatten out as it spins faster, leading to changes in its path.

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What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 11 REarth before coming to rest momentarily?

Answers

the launch speed of the projectile is approximately 12.3 km/s that rises vertically above the Earth to an altitude equal to 11 REarth before coming to rest momentarily

We can use the conservation of energy principle to solve this problem. At the highest point, the projectile has zero kinetic energy and a certain potential energy, given by:

PE = mgh

where m is the mass of the projectile, g is the acceleration due to gravity, and h is the altitude above the Earth's surface. The potential energy can also be expressed in terms of the gravitational potential energy per unit mass, which is given by:

U = GM/r

where M is the mass of the Earth, r is the distance from the center of the Earth to the projectile, and G is the gravitational constant. At the highest point, the projectile has a distance of 12 REarth from the center of the Earth (11 REarth above the surface plus the radius of the Earth).

Setting these two expressions for potential energy equal to each other, we have:

mgh = GMm/r

where we can cancel the mass m from both sides to obtain:

gh = GM/r

Solving for the speed v at the launch point, we have:

v = sqrt(2gh)

where we can substitute:

h = 11 REarth

r = 12 REarth

g = GM/(12 REarth)2

Substituting these values and using the fact that REarth = 6.37×106 m, we get:

v = sqrt(2×(11+1)REarth×GM/(12REarth)) = 12.3 km/s

where we used the fact that the total distance traveled by the projectile is 2 REarth, so the factor of (11+1) appears in the expression for h.

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A charge +q and a charge +q are placed at opposite corners of a square. Will a third point charge experience a greater force if it is placed at one of the empty corners of the square, or at the centre of the square? Explain.

Answers

A three - point bending charge will not encounter any force in the square's centre because the electric field there is zero.

when a +q point charge is applied to a square's corner?

An applied electric strength about 2 N/C is seen at the centre of a square whenever a point charges of +q is applied to one of its corners. Assume the three remaining corners of a square receive three identical +q charges.

when the 2-point charges +q & Q are positioned opposite to one another?

Two point charges +q and -q are set at a location on the unirom ring that is diameterically opposite, with -q at the bottom and in contact with an inclination plane of a perfect insulator. In the presence of a uniform vertical electric field, a total mass of m remains in equilibrium just on rough inclined plane.

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1. The cylinder moves at this velocity all the way around the circular path. Yet when we view the side to side motion, the cylinder does not appear to move with a constant velocity. At which location during the side-to-side motion does the cylinder appear to have the minimum side-to-side velocity? 2. At which location during the side-to-side motion does the cylinder appear to have the maximum side-to-side speed

Answers

As a result, at these locations, the cylinder's side-to-side velocity looks to be the slowest. Therefore, it appears that the cylinder has the highest side-to-side motion at these locations.

What happens when a body moves uniformly in a circular manner along a path?

An acceleration towards the circular path's middle is felt by a body travelling along a circular path at constant speed. A centripetal force produces the acceleration, which is known as centripetal acceleration.

Does a body travelling in a uniform circle have a constant speed? If not, why not?

To put it simply, an object moving in a circle at a consistent speed is said to be in uniform circular motion. Although the object is moving at a constant pace, its velocity is changing. Since velocity is a vector,

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A 150-g apple is falling from a tree. What is the impulse that Earth exerts on it during the first 0.50m of its fall? The next 0.50 m ?

Answers

The impulse that Earth exerts on the apple during the next 0.50m of its fall is:

Δp = m * v2 - m * v1 = 0.15 kg * 4.43 m/s - 0.15 kg * 3.13 m/s = 0.16 Ns.

The impulse that Earth exerts on an object is equal to the change in momentum of the object. The momentum of an object is given by:

p = m * v

where p is momentum, m is mass, and v is velocity.

During free fall, the velocity of the object increases due to the acceleration due to gravity, which is approximately 9.81 m/s^2. Therefore, the velocity of the apple after falling a distance of d is:

v = sqrt(2gd)

where g is the acceleration due to gravity and d is the distance fallen.

The change in momentum of the apple during the first 0.50m of its fall is:

Δp = p2 - p1 = m * v2 - m * v1

where p1 is the initial momentum, p2 is the final momentum, v1 is the initial velocity (which is 0), and v2 is the final velocity after falling a distance of 0.50m.

Plugging in the values:

m = 150 g = 0.15 kg

g = 9.81 m/[tex]s^2[/tex]

d = 0.50 m

The final velocity of the apple after falling 0.50m is:

v2 = sqrt(2gd) = sqrt(29.810.50) = 3.13 m/s

Therefore, the impulse that Earth exerts on the apple during the first 0.50m of its fall is:

Δp = m * v2 - m * v1 = 0.15 kg * 3.13 m/s - 0.15 kg * 0 m/s = 0.47 Ns

For the next 0.50m of its fall, the initial velocity is now 3.13 m/s and the final velocity after falling another 0.50m is:

v2 = sqrt(2gd) = sqrt(29.811.00) = 4.43 m/s.

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