A man walking at 3.56 m/s accelerates at 2.50 m/s2 for 9.28 s. How far does he get?

Answers

Answer 1

The man who walks at 3.56 m/s and accelerates at 2.50 m/s2 for 9.28 s would walk a distance of  135.245 meters.

Kinematic motion

We can use the kinematic equation:

distance = initial velocity x time + (1/2) x acceleration x time^2

To use this equation, we need to find the initial velocity of the man before he started accelerating. We can do this using the formula:

final velocity = initial velocity + acceleration x time

At the start, the man's velocity was 3.56 m/s, and he accelerates at 2.50 m/s^2 for 9.28 s. Therefore, his final velocity can be calculated as:

final velocity = 3.56 + 2.50 x 9.28

final velocity = 26.08 m/s

Now we can use the distance formula:

distance = initial velocity x time + (1/2) x acceleration x time^2

with initial velocity being 3.56 m/s, time being 9.28 s, acceleration being 2.50 m/s^2, and final velocity being 26.08 m/s:

distance = 3.56 x 9.28 + (1/2) x 2.50 x (9.28)^2

distance = 32.968 + 102.277

distance = 135.245 m

Therefore, the man traveled a distance of approximately 135.245 meters.

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Related Questions

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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(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.

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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A 40kg child on a 50kg bike is riding with a velocity of 10 m/s to the northern.

Answers

Answer:

Explanation:

The vertex of an absolute function is the minimum or the maximum point of the graph

The graph that could be  is (a) an absolute value graph has a vertex at (2, 1)

The function is given as:

And the coordinates of the vertex (h,k) are said to be positive.

From the list of given options, only the first option has both coordinates of the vertex to be positive i.e. (2,1)

Hence, the graph that could be  is (a)

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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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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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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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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

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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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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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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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.

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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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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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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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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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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Sabrina takes a self-report, objective personality rest that asks her to agree or disagree with over 300 statements such as, "I forgive people easily" and "I often got into trouble when I younger". Sabrina is probably taking the

Answers

Sabrina is probably taking the Big Five Personality Test.

What are the three impartial personality assessments?

The Beck Depression Inventory, Millon Clinical Multiaxial Inventory-III, Minnesota Multiphasic Personality Assessment, and Child Behavior Checklist are some of the frequently used objective personality tests.

What are a few illustrations of impartial personality tests?

Examples of objective personality assessments include the Big Five Personality Inventory, the Millon Clinical Multiaxial Inventory (MCMI), and the Minnesota Multiphasic Personality Inventory (MMPI).

What is a psychological assessment that is objective and measures mental disorders?

The Minnesota Multi-Phasic Personality Inventory (MMPI), which rates people on personality, psychological, and emotional factors, is arguably the most popular of the objective personality tests.

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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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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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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.

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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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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1.datemine the mechanical energy of mass 200g of bird fling at 12m/s at height of some use (g=12m/s²)
2. calculate the GPF of the following
A. A 20kg lifed 2m into air
B. sixty 6kg boxes lifted into shelf 2m high
3. calculate the kinetic energy of
A. 3g of bullest travelling at 40m/s
B. A car of mass 1200kg that travelling 60m in 3 seconds ​

Answers

The mechanical energy of the bird is 26.4 J.

Mechanical energy calculation.

To determine the mechanical energy of the bird, we need to calculate its kinetic energy and potential energy.

Kinetic energy = 1/2 * mass * velocity^2

Potential energy = mass * gravity * height

Given:

Mass = 0.2 kg

Velocity = 12 m/s

Height = unknown

Gravity = 12 m/s^2

Kinetic energy = 1/2 * 0.2 * 12^2 = 14.4 J

To find the potential energy, we need to know the height. Let's assume it is 5 meters.

Potential energy = 0.2 * 12 * 5 = 12 J

Mechanical energy = kinetic energy + potential energy

Mechanical energy = 14.4 + 12 = 26.4 J

Therefore, the mechanical energy of the bird is 26.4 J.

A.

Gravitational potential energy = mass * gravity * height

GPE = 20 * 9.81 * 2 = 392.4 J

The work done to lift the object is equal to its gravitational potential energy, so the GPF (Gravitational Potential Force) is 392.4 N.

B.

Total mass of boxes = 60 * 6 = 360 kg

Gravitational potential energy = mass * gravity * height

GPE = 360 * 9.81 * 2 = 7069.2 J

The work done to lift the boxes is equal to their gravitational potential energy, so the GPF is 7069.2 N.

Therefore, the GPF for the 20 kg lift is 392.4 N and the GPF for the 60 boxes is 7069.2 N.

A.

Kinetic energy = 1/2 * mass * velocity^2

KE = 1/2 * 0.003 * 40^2 = 2.4 J

Therefore, the kinetic energy of the 3 g bullet travelling at 40 m/s is 2.4 J.

B.

We can calculate the average velocity of the car using the formula:

Average velocity = distance / time

Average velocity = 60 / 3 = 20 m/s

Kinetic energy = 1/2 * mass * velocity^2

KE = 1/2 * 1200 * 20^2 = 240,000 J

Therefore, the kinetic energy of the car is 240,000 J.

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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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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

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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.

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

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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I need help answering the following question. Thank you!

Answers

a. The graph does not appear to be symmetrical.

b. The graph appears to be bimodal, as there are two distinct peaks.

c. The data appear to be positively skewed, as the tail of the graph extends further to the right than to the left.

d. There appear to be outliers for values less than 3.5 and greater than 6.75, as these values are relatively far away from the rest of the data and do not fit the overall pattern of the graph.

What is a graph?

A graph is described as a structure amounting to a set of objects in which some pairs of the objects are in some sense "related.

Graphs are a popular tool for graphically illuminating data relationships.

In conclusion,  A graph serves the purpose of presenting data that are either too numerous or complex to be properly described in the text while taking up less room.

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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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