negative work means a. the kinetic energy of the object increases. b. nothing; there is no such thing as negative work. c. the object has negative velocity. d. the kinetic energy of the object decreases. e. the object does not move.

We are given that a sphere Z moves with a relative speed of 1 m/s with respect to sphere X which moves at 5 m/s. Let's draw the velocity vectors of the spheres:

Where:

From the relative velocity equation we have:

Where:

Since we are given the relative velocity we can plug in the values to get the velocity of "z":

Now, we do the same but now using the sphere Y:

Now, we use the relative velocity equation for these velocities:

In this case, we have that:

Now, we subtract the velocity of "y" from both sides:

Substituting the values:

Solving the operations:

Therefore, the velocity of "Z" relative to "Y" is 4 m/s.

To get the resonant frequencies of a single cohesive wave, just use formula v = f.The wave velocity is denoted by the letter "v," while the wave's distance is denoted by the symbol "."

The length of the string:

L=75.0 cm=0.75 m

The two resonant frequencies:

450 Hz Λ 308 Hz

a) f1=142 Hzf1​=142 Hz

b)v=213 msv=213 sm​

Gamma rays often have the greatest frequency, while radio waves typically have the lowest.

The fundamental frequency is the lowest frequency that can be produced by a certain instrument.A first harmonic of a instrument is another name for the fundamental frequency.

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Answer:

A) Velocity

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Answer:

A) Velocity

Explanation:

They Have both a magnitude and direction ..

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The spring is compressed by 0.84 meters.

The potential energy of the block at the starting point is given by,

mgh = 2 x 9.8 x h, where h is the height of the starting point above the bottom of the incline.

Using the given inclination angle of 37 degrees, we can find h as,

h = 4.8 sin(37) = 2.88 m

We can equate the potential energy at the starting point to the energy stored in the compressed spring and the kinetic energy at the bottom of the incline,

mgh = 1/2 kx^2 + 1/2 mv^2

where x is the distance that the spring is compressed, v is the velocity of the block at the bottom of the incline, and k is the spring constant.

Since the block is sliding without friction, we can use conservation of energy to find v as,

mgh = 1/2 mv^2

v^2 = 2gh

v = √(2gh)

Substituting this expression for v in the earlier equation,

mgh = 1/2 kx^2 + 1/2 m(2gh)

2gh = kx^2/m + 2gh

x^2 = 2ghm/k

x = √(2ghm/k)

Substituting the known values,

x = √(2 x 9.8 x 2 x 2.88 / 1000) = 0.84 m

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--The complete question is, A spring is fixed at the bottom end of an incline of inclination 37∘. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away (along the inclined plane) from the spring. The spring constant is 1000 N/m. How far is the spring compressed?--

The harmonics are sometimes referred to as the overtone series.

The frequencies that are capable of forming standing waves on a string attached at both ends from the fundamental frequency on up are called harmonics. In music, harmonics are overtones that accompany a sound wave's fundamental frequency.

These waves are at frequencies that are whole number multiples of the fundamental frequency (f₀).

These waves are at frequencies that are whole number multiples of the fundamental frequency (f₀).

2f₀ is twice f₀, 3f₀ is three times f₀, and so on.

These harmonics exist in a series of discrete frequencies that are proportional to the frequency of the fundamental pitch. Therefore, the harmonics are sometimes referred to as the overtone series.

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The equation for finding the net torque on a dipole is given by τ = p × E, where τ represents the torque, p is the dipole moment, and E is the electric field.

A dipole exhibits a torque because it consists of two opposite charges separated by a distance, and when placed in an electric field, these charges experience forces in opposite directions, causing a rotational effect or torque. The torque tends to align the dipole moment with the direction of the electric field, and the strength of the torque depends on the magnitude of the dipole moment and the electric field, as well as the angle between them.

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In order the weight feels tripled, the elevator should be moving upward with acceleration 2g m/s² or two times of gravitational acceleration.

The free body diagram is shown on the attached picture.

The scale will read the normal force.

Suppose the elevator is moving upward. Using the Newton's second law of motion:

∑F = m.a

N - w = ma

N = mg + ma

Suppose the elevator is moving downward.

w - N = ma

N = w - ma

N = mg - ma

Where:

m = mass

a = acceleration

g = gravitational acceleration

From the above 2 cases, in order the scale to read the measured weight higher than its actual weight, N should be greater than w, or N = mg + ma, or the elevator should be moving upward.

The weight feels tripled means N = 3 mg.

Hence,

3 mg = mg + ma

a = 3g - g = 2g

Therefore, in order the weight feels tripled, the elevator should be moving upward with acceleration 2g m/s²

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Answer:

change  in velocity over change in time

Explanation:

The charge of a newly formed oxygen ion would be -2

Hope this helps :)

Answer:

It is due to the large impulse is imparted on the flour.

Explanation:

A ball is moving faster.

When a ball is moving faster strikes to the flour, the change in momentum is large and thus the impulse imparted on the flour is large.

Impulse = change in momentum

So, as the flour experiences large impulse and large momentum so that the flour spreads out.

If the change in momentum is large so the flour spreads out is more.  

When a ball moves with faster velocity, then a greater impulse force is generated that causes more flour to spread out.​

Momentum can be defined as the mass in motion. It is the product of mass and velocity. It is given by the formula,

p = mv

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

As we know that the ball has some mass and the ball is moving with some velocity, therefore, the ball will have some momentum. Also, we know that when the ball will hit the flour the momentum of the ball will become zero, therefore, the change in momentum in such a short time will generate an impulse.

Thus, if the velocity of the ball is more than the impulse generated will be more which will cause more flour to spread out.​

Hence, When a ball moves with faster velocity, then a greater impulse force is generated that causes more flour to spread out.​

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Answer:

Explanation:

\(E=\frac{q}{F}\) where q is the charge of the electron and F is the electrostatic force. Filling in:

\(E=\frac{-1.6*10^{-19}}{9.2*10^{-15}}\) which gives you an electric field magnitude of

E = -1.7 × \(10^{-5\) C/N

Answer:

it's zero

Explanation:

it is there is your answer

(A) The pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.
(B) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow

(a) The pressure transients at the mid-point of the pipeline can be calculated using the water hammer equation. Water hammer refers to the sudden changes in pressure and flow rate that occur when there are rapid variations in fluid flow. The equation is given by:

ΔP = (ρ × ΔV × c) / A

Where:

ΔP = Pressure change

ρ = Density of water

ΔV = Change in velocity

c = Wave speed

A = Cross-sectional area of the pipe

First, let's calculate the change in velocity:

ΔV = Q / A

Q = Flow rate = 0.12 m/s

A = π × ((d/2)^2 - ((d-2t)/2)^2)

d = Internal diameter of the pipe = 300 mm = 0.3 m

t = Pipe wall thickness = 5 mm = 0.005 m

Substituting the values:

A = π × ((0.3/2)^2 - ((0.3-2(0.005))/2)^2

A = π × (0.15^2 - 0.1495^2) = 0.0707 m^2

ΔV = 0.12 / 0.0707 = 1.696 m/s

Next, let's calculate the wave speed:

c = √(E / ρ)

E = Young's modulus of steel = 210x10^9 Pa

ρ = Density of water = 1000 kg/m^3

c = √(210x10^9 / 1000) = 4585.9 m/s

Finally, substituting the values into the water hammer equation:

ΔP = (1000 × 1.696 × 4585.9) / 0.0707 = 1,208,277 Pa

Therefore, the pressure transients at the mid-point of the pipeline are approximately 1,208,277 Pa.

(b) Friction in the pipeline affects the calculated pressure transients by increasing the overall resistance to flow. As water moves through the pipe, it encounters frictional forces between the water and the pipe wall. This friction causes a pressure drop along the length of the pipeline.

The presence of friction results in a higher effective wave speed, which affects the calculation of pressure transients. The actual wave speed in the presence of friction can be higher than the wave speed calculated using the Young's modulus of steel alone. This higher effective wave speed leads to a reduced pressure rise during the transient event.

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The moon has a gravitational field strength one-sixth that of earth's. If a person has a mass of 60kg on Earth. Weight on moon = 10kg.

The weight of an object is the force exerted on it by gravity, which is proportional to its mass and the strength of the gravitational field. The gravitational field strength on the moon is one-sixth that of Earth's, meaning that the force of gravity pulling on objects is weaker on the moon.

To calculate the weight of a person on the moon, we need to multiply their mass by the gravitational field strength of the moon, which is 1/6 or 0.1667 of the Earth's. Therefore, the weight of a 60kg person on the moon would be:

Weight on moon = mass x gravitational field strength on moon

Weight on moon = 60kg x 0.1667

Weight on moon = 10kg

The person would weigh 10kg on the moon, which is significantly less than their weight on Earth. This is because the gravitational pull on the moon is much weaker than that on Earth. It's important to note that the person's mass remains the same regardless of where they are, while their weight varies depending on the strength of the gravitational field.

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Most Kenyan athletes do their practice in mountainous areas because it helps in developing their physical abilities and strengthening their lungs. It is believed that the altitude and tough terrain conditions make it possible to run more efficiently at lower altitudes. Training at high altitudes has several benefits.

First, the high-altitude environment has less oxygen, which means that athletes have to work harder to breathe and produce energy. This helps in developing their physical abilities and strengthening their lungs. As the body gets used to this environment, it is better equipped to handle oxygen and energy more efficiently, which can lead to improved performance. Second, training at high altitudes can lead to an increase in the number of red blood cells in the body. Red blood cells are responsible for carrying oxygen to the muscles, and an increase in their numbers can improve an athlete's endurance levels. Third, high-altitude training can also help in weight loss as it burns more calories than training at lower altitudes. This can help athletes maintain their weight and improve their fitness levels. Fourth, the tough terrain conditions in mountainous areas can also help athletes in developing their skills.

Running on uneven surfaces can help improve balance, agility, and coordination, which are essential for a runner. Training at high altitudes also has some drawbacks. The most common one is altitude sickness, which is caused by the lack of oxygen in the air. Altitude sickness can cause headaches, nausea, vomiting, and shortness of breath. It can also lead to dehydration, which can further worsen the symptoms. Additionally, high-altitude training can be expensive as athletes need to travel to mountainous areas to train. The cost of living and training in such areas can be high, which can be a barrier for some athletes.

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Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 18 × 1.62

We have the final answer as

Hope this helps you

Explanation:

Let A and B be two objects of mass 'm' each

let A have volume V1 and B have volume V2 where V1 > V2

we know that density = mass/volume

here density is inversely proportional to volume

here density of A = m/V1

and density of B = m/V2

since V1 > V2

on comparing,

m/V1 < m/V2

i.e., density of B > density of A

Answer:

please find attached file

Explanation:

Answer:

Difficult to Pour.

Explanation:

Answer:

thank you very much for giving me points.

Explanation:

I really appreciate it so muchhhhh so muchhhh thanks again

(a) The radius of the wheel = 22 ft

The wheel is rotating 13 RPM counterclockwise

Angular speed = angular velocity = ω = 2πf = 2 × π × 13 = 26π rad/min (since 1 rev = 2π radians)

Since 1 min = 60 sec, ω = (26π)/60 rad/sec = 13π/30 rad/sec

The linear speed v of a point on the rim of the wheel is given by v = r × ω = 22 × 13π/30 = 22.82 ft/s

Therefore, w = 13π/30 rad/sec and v = 22.82 ft/sec

(b) The radius of the wheel = 6 inThe wheel is rotating at 30°/sec

The angular speed = ω = 30°/sec × (π/180°) = π/6 rad/sec

The linear speed v of a point on the rim of the wheel is given by v = r × ω = 6 × π/6 = π in/sec

The angular speed in RPM can be calculated as follows:

In 1 min, the angle rotated = 360°No. of seconds in 1 min = 60∴

The angle rotated in 1 sec = 360°/60 = 6° or (π/30) radThe angular speed in rad/sec and the angular speed in RPM is given by w = π/6 rad/sec and w = (30/π) × π/6 = 5 RPM

(c) The radius of the Earth = 3960 milesThe circumference of the Earth = 2 × π × radius = 2 × π × 3960 ≈ 24,902 miles (approx.)

One rotation of the Earth is completed in 24 hours or 24 × 60 × 60 = 86,400 secLinear speed v of a point on the equator of the Earth is given byv = circumference of the Earth/time taken for 1 rotation= 24,902/86,400 ≈ 0.2887 miles/sec

Therefore, v = 0.2887 × 60 × 60 = 1040 miles/hourAngular speed = ω = 2πf = 2π/Twhere T = time taken for 1 rotation of the Earth= 24 hours = 24 × 60 × 60 = 86,400 sec∴ ω = 2π/86,400 rad/sec

Angular speed in RPM can be calculated as follows:In 1 min, the angle rotated = 360°No. of seconds in 1 min = 60∴

The angle rotated in 1 sec = 360°/60 = 6° or (π/30) radThe angle rotated in 24 hours = 360°No. of seconds in 24 hours = 24 × 60 × 60 = 86,400∴

The angle rotated in 1 sec = 360°/86,400 = 1/240° or π/43,200 radThe angular speed in RPM is given by w = (360°/43,200) × 60 = 0.1666 RPM (approx.)

(d) The radius of the tire = 12 inchesThe speed of the car = 75 mphLet the car travel for 1 hour in which the tire makes x revolutions∴

The distance travelled by the car in 1 hour = 75 miles = circumference of the tire × x= 2π × 12 × x inches= 24πx inchesTherefore, 24πx = 75 × 5280 × 12 inches or x = 19600 revolutions∴

The tire makes 19600 revolutions in 1 hour= 19600 × 2π radians= 39200π radians∴ Angular speed = ω = 39200π/3600 = 109.11 rad/sec= (109.11/2π) RPM= 17.36 RPM (approx.)

Therefore, the angular speed in rad/sec is 109.11 rad/sec and the angular speed in RPM is 17.36 RPM.

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First thing's first, we need to find the Net Force of the system in order to solve for acceleration. Gravity and Normal Force cancel out, so we don't really have to worry about anything on the y-axis. However, looking at the x-axis, we see two forces: the Force of the push and Frictional Force.

This means that:

\(F_{net}\) = 500 - 100

\(F_{net}\) = 400 (N)

Recall that Force = Mass * Acceleration. Since we have the Force and Mass, we can easily solve for Accel:

F = ma

a = \(\frac{F}{m}\)

Plug the net force and mass into the equation:  

a = \(\frac{F}{m}\)

a = \(\frac{400}{50}\)

a = 8 (m/\(s^{2}\))

The box's acceleration is 8 m/\(s^{2}\)

If you have any questions about how I got to the answer, just ask :)

Answer:

Reason for the difference in the ranges of the ball and the cart:

"the average speed of the cart is less than the instant speed of the cart at the time of throwing the ball".

Explanation:

Assuming that the air friction is negligible.

Given that the ball lands on the ground a little before the cart.

So, the range of the ball is more than the range of the cart in the same time interval.

Let the instant speed of the cart is v m/s  at the time of throwing the ball in the vertically upward direction, so the speed of the ball in the horizontal direction = v m/s.

Let t be the total time of flight of the ball.

As the gravitational force is acting in the downward direction so it will not change the horizontal velocity of the ball.

So, the range covered by the ball \(= vt\) m.

Let u be the average speed of the cart for the time t s (same as the time of flight of ball).

So, the distance by the cart \(= ut\) m

As the range of the ball is more than the range of the cart in the same time interval, so

\(vt > ut \\\\\Rightarrow v>u.\)

So, the reason for the difference in the ranges of the ball and the cart is "the average speed of the cart is less than the instant speed of the cart at the time of throwing the ball".

The eye forms the image on the retina. The correct option is c.

When light enters the eye, it passes through the cornea, the clear outer covering of the eye, and then through the pupil, which is the opening in the center of the iris. The iris is the colored part of the eye that helps to control the amount of light that enters. The lens then focuses the light onto the retina, which is a layer of light-sensitive cells located at the back of the eye.

These cells, called photoreceptors, convert the light into electrical signals that are sent to the brain via the optic nerve. The brain then processes these signals to create the visual image that we perceive. Therefore, the retina is where the actual image is formed in the eye, and the optic nerve carries this information to the brain for interpretation. Therefore, the correct option is c.retina.

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Answer:

True

Explanation:

mass, length, time, electric current, temperature, luminous intensity and amount of substance

The limit resistance of the footing is determined to be [insert numerical value] now.

The limit resistance of a footing refers to its ability to resist the maximum load or force it can withstand before failure or excessive settlement occurs. In this case, considering that the footing is embedded 0.6m in the ground (d = 0.6m), we can calculate the limit resistance using relevant engineering principles.

The limit resistance of a footing is influenced by various factors, including the type of soil, the dimensions of the footing, and the depth at which it is embedded. When a footing is embedded deeper into the ground, it benefits from the increased bearing capacity provided by the underlying soil layers.

By embedding the footing 0.6m into the ground, it effectively increases the load-bearing capacity compared to a footing that sits on the ground surface. The additional depth allows the footing to interact with deeper, more compacted soil layers that can provide greater resistance to vertical loads.

To determine the limit resistance of the footing, it is necessary to perform geotechnical calculations and consider factors such as the ultimate bearing capacity of the soil and the size and shape of the footing. These calculations typically involve considering the soil properties, such as its shear strength and cohesion, along with the applied load and the depth of the footing.

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The average acceleration between t = 5.6 s and t = 8.5 s is 2.31  m/s²

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.

Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds

The acceleration is given by

a = (-1 - 5.7)/ 2.9

a = -  2.31 m/s²

|a | = 2.31 m/s²

Thus, the object's acceleration is 2.31 m/s²

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The speed of the cart when it has been pushed 8 m with a constant 8 N horizontal force is approximately 2.6 m/s.

To find the speed of the cart when it has been pushed 8 m with a constant 8 N horizontal force, we can use the principles of Newton's laws of motion.

The force applied to the cart is 8 N, and the mass of the cart is 19 kg. We can use Newton's second law, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass

F = m * a

Where F is the net force, m is the mass, and a is the acceleration.

In this case, the net force is 8 N, and the mass is 19 kg. We can rearrange the equation to solve for acceleration

a = F / m

a = 8 N / 19 kg

a = 0.421 N/kg

Now, we can use the kinematic equation that relates distance (d), initial velocity (v₀), acceleration (a), and final velocity (v)

v² = v₀² + 2 * a * d

Since the cart is initially at rest (v₀ = 0), the equation simplifies to

v² = 2 * a * d

Substituting the values, we get

v² = 2 * 0.421 N/kg * 8 m

v² = 6.736 m²/s²

Taking the square root of both sides to find the speed (v), we get

v = √6.736 m/s

v = 2.6 m/s

Therefore, the speed of the cart when it has been pushed 8 m with a constant 8 N horizontal force is approximately 2.6 m/s.

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Answer:

im going to guess the 3rd one

Explanation: