Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take before the arrow hits the ground below (step 1 of a quesiton will need this answer for a future question)
a 2 seconds
b. 1 second
c.4 seconds
d 19.6​

The time of the rock in flight is 10.71 s.

This is the total time taken for an object or a projectile to return back to the same plane at which it was projected.

To calculate the time of the rock in flight, we use the formula below.

Formula:

Where:

make t the subject of the equation

t = (v-u)/g.................... Equation 2

From the question,

Given:

Substitute these values into equation 2

Hence, the time of the rock in flight is 10.71 s.

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The rock was in the flight for 11.7 s

This is defined as the total time spent by an object in air.

From the question given above, the following data were obtained:

v² = u² + 2gh

115² = 10² + (2 × 9.8 × h)

13225 = 100 + 19.6h

Collect like terms

13225 – 100 = 19.6h

13125 = 19.6h

Divide both side by 19.6

h = 13125 / 19.6

h = 669.64 m

h = ½gt²

669.64 = ½ × 9.8 × t²

669.64 = 4.9 × t²

Divide both side by 4.9

t² = 669.64 / 4.9

Take the square root of both side

t = √(669.64 / 4.9)

t = 11.7 s

Thus, the rock was in the flight for 11.7 s

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The time required for half of the oil droplets to coalesce is 34.65 minutes.b) Calculation to find how long it takes for practically all of the oil droplets to coalesce:To find the time it would take for practically all of the oil droplets to coalesce, we need to use the following formula and solve for time when n is equal to 0.0 = e^(-0.02t)-infinity = -0.02tNo oil droplets remain after an infinite amount of time. Therefore, it takes an infinite amount of time for all the oil droplets to coalesce.Answer: It takes an infinite amount of time for all the oil droplets to coalesce.

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Answer:

B. Ritcher

Explanation:

Answer:

b

Explanation:

Answer:

486,750 kg*m/s

Explanation:

Momentum is mass*velocity

M = m*v

M = 8850kg*55m/s

M = 486,750 kg*m/s

The momentum of an 8850 kg medium truck that is traveling with a velocity of 55 m/s west on the highway is 486750 kg m / s.

Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force acting on the particle equals the time rate of change of momentum.

According to Newton's second law, if a particle is subjected to a constant force for a specific amount of time, the result of the force and time (referred to as the impulse) is equal to the change in momentum.

Given:

The mass of the truck is, m = 8850 kg,

The velocity of the truck is, v = 55 m/s,

Calculate the momentum of the truck as shown below,

Momentum = m × v

Momentum = 8850 × 55

Momentum = 486750 kg m / s

Thus, the Momentum of the truck is 486750 kg m / s.

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The planet that has ⅒ of the earth's gravity is the moon.

Gravity is the force exerted by any object with mass on any other object with mass.

Gravity is the force on Earth's surface, of the attraction by the Earth's masses, and the centrifugal pseudo-force caused by the Earth's rotation, resulting from gravitation.

The gravity on the planet Earth is 1 with a acceleration due to gravity of 9.8m/s². One-tenth of this is 0.1 (0.98m/s²).

The planet with the above gravity is the moon with a gravity of 0.166.

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Answer:

  v = \(n \frac{\hbar }{m r}\)

the sppedof the electron is also quantized

Explanation:

The angular momentum of a rotating body is

         L = m v r

in Bohr's atomic model the quantization postulate is that the angular momentum is equal to

         L = n \(\hbar\)

we substitute

        n \(\hbar\) = m v r

        v = \(n \frac{\hbar }{m r}\)

where n is an integer.

Therefore, the sppedof the electron is also quantized, that is, sol has some discrete values.

B. Noise decreases the loudness of analog signals but does not affect the loudness of digital signals.

Noise is any extraneous interference that can affect the quality of a signal being transmitted.

Noise can cause analog signals to become weaker or distorted, which can reduce the loudness of the signal and make it more difficult to interpret. In contrast, noise does not affect the loudness of digital signals in the same way.

Digital signals are transmitted using a series of ones and zeros, which can be easily distinguished from noise. This means that digital signals are less susceptible to the effects of noise, and are generally considered to be more reliable than analog signals for this reason.

Other reasons to use digital signals instead of analog signals include the fact that digital signals are easier to store and transmit over long distances, and that they can be more easily processed by computers.

\(\bold{ \: \purple{Hope \: This \: Helps \: You!}}\)

B. Noise decreases the loudness of analog signals but does not affect the loudness of digital signals.

Noise is any extraneous interference that can affect the quality of a signal being transmitted. Noise can cause analog signals to become weaker or distorted, which can reduce the loudness of the signal and make it more difficult to interpret. In contrast, noise does not affect the loudness of digital signals in the same way.

Digital signals are transmitted using a series of ones and zeros, which can be easily distinguished from noise. This means that digital signals are less susceptible to the effects of noise, and are generally considered to be more reliable than analog signals for this reason.

Other reasons to use digital signals instead of analog signals include the fact that digital signals are easier to store and transmit over long distances, and that they can be more easily processed by computers.

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Answer: Displacement = -2.39 m

Explanation: The displacement of an object is the difference between its final position and its initial position. In this case, the object moved from x = -1.88 m to x = -4.27 m, so its displacement is:

Displacement = Final position - Initial position

Displacement = (-4.27 m) - (-1.88 m)

Displacement = -4.27 m + 1.88 m

Displacement = -2.39 m

Therefore, the displacement of the object is -2.39 m.

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The amount of energy lost to air friction is 3414.5 J.

We can use the principle of conservation of energy, which states that the total energy of a system remains constant unless acted upon by external forces. At the top of the building, the ball has potential energy due to its height above the ground. As it falls, this potential energy is converted to kinetic energy, which is then dissipated as heat due to air friction when the ball strikes the ground.

The potential energy of the ball at the top of the building is given by:

PE = mgh

Where

Substituting the given values, we get:

PE = (5 kg)(9.81 m/s^2)(90 m) = 4414.5 J

The kinetic energy of the ball just before it strikes the ground is given by:

KE = (1/2)mv^2

Where v is the velocity of the ball just before it strikes the ground. Substituting the given value, we get:

KE = (1/2)(5 kg)(20 m/s)^2 = 1000 J

The energy lost to air friction is the difference between the initial potential energy and the final kinetic energy:

Energy lost = PE - KE = 4414.5 J - 1000 J = 3414.5 J

Therefore, the amount of energy lost to air friction is 3414.5 J.

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Table A agrees with the information in the problem. After 25.0 seconds, the velocity of the plane is 187 m/s.

Based on the given problem, we need to determine the velocity of a plane after 25.0 seconds. The plane is initially flying east at a velocity of 115 m/s, and it experiences an acceleration of 2.88 m/s² in the northwest direction.

Let's analyze each option and calculate the final velocity (Vf) of the plane after 25.0 seconds:

Option A:

Vi = 115 m/s

a = 2.88 m/s²

t = 25.0 s

Using the equation Vf = Vi + at, we can calculate:

Vf = 115 m/s + (2.88 m/s²)(25.0 s) = 115 m/s + 72 m/s = 187 m/s

Option B:

V₁ = 2.04 m/s

a = 2.04 m/s²

t = 25.0 s

Using the equation Vf = V₁ + at, we can calculate:

Vf = 2.04 m/s + (2.04 m/s²)(25.0 s) = 2.04 m/s + 51 m/s = 53.04 m/s

Option C:

V₁ = 115 m/s

a = -2.04 m/s²

t = 25.0 s

Using the equation Vf = V₁ + at, we can calculate:

Vf = 115 m/s + (-2.04 m/s²)(25.0 s) = 115 m/s - 51 m/s = 64 m/s

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The work done on the crate to move it vertically is 7000 J.

The work done to move a 1400-N crate at constant speed 5.0 m vertically is 7000 J. Here's the explanation:When an object is lifted up at a constant speed, the force applied to it is equal and opposite to the force of gravity acting on it. As a result, no net force is applied to the object, and no work is performed against it.The formula to determine the work done isW = FdwhereW is work doneF is forceand d is distance coveredTo calculate the work done on the crate to move it vertically upwards, we need to determine the force that needs to be applied to overcome gravity acting downwards. We know that force is equal to mass multiplied by acceleration, which means that the force required to overcome gravity acting on the crate is given by:F = mgwhereF is the force in Newtonsm is the mass in kilogram is the due to acceleration, 9.8 m/s²Substitute the given values into the formula:F = 1400 NNext, we need to determine the distance over which the crate is lifted vertically, which is given as d = 5.0 m.Substitute the given values into the formula: d = 5.0 mFinally, substitute the values into the formula to get the work done.W = FdW = 1400 N × 5.0 mW = 7000 J.

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The amount of work required to move the crate is determined as 7,000 J.

The amount of work required to move the crate is calculated from the product of applied force and the displacement of the crate.

Mathematically, the formula for work done is given as;

W = Fd

where;

The amount of work required to move the crate is calculated as;

F = 1400 N x 5.0 m

F = 7,000 J

Thus, the amount of work required to move the crate is calculated from the product of applied force and the displacement of the crate.

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Answer: B. DLC

B. Visual Updates

B. Competition and expectations

D. Every three months

D. Game expansion packs

C. Visual updates

A. Game expansion

C. Concept art books

A. Create DLC

A. Check licensing agreements

D. Separate her personal finances from the business finances

B. Soundtrack

C. Group dynamics

C. To open a business bank account

B. Onboarding activities

Explanation: :)

The wavelength of light that should be used to obtain fringes that are 0.0045 m wide after reducing the distance between the screen and the slit by half is 2.25 * 10^7 Å.

Huygens's Principle states that every point on a wavefront can be considered as a source of secondary spherical wavelets that spread out in all directions with the same speed as the original wave. The new wavefront is formed by the envelope of these secondary wavelets at a later time.

Now, let's consider a Young's double-slit experiment. In this experiment, when light passes through two narrow slits, it creates an interference pattern on a screen behind the slits. The fringe width is the distance between two consecutive bright or dark fringes in the pattern.

Given that the fringe width obtained is 0.6 cm and the wavelength of light used is 4500 Å (Angstroms), we can calculate the wavelength of light required to obtain fringes that are 0.0045 m wide.

We can use the formula for fringe width in Young's double-slit experiment:

w = (λ * D) / d

Where:

w is the fringe width,

λ is the wavelength of light,

D is the distance between the screen and the double slits, and

d is the distance between the two slits.

Let's calculate the value of D/d using the given information:

D/d = w / λ

= 0.006 m / 4500 Å (1 m = 10^10 Å)

= 0.006 * 10^10 / 4500 m^-1

Now, if the distance between the screen and the slit is reduced by half, the new value of D/d would be:

(D'/d) = (0.006/2) * 10^10 / 4500 m^-1

Now, we can rearrange the equation to solve for the new wavelength (λ'):

(λ' * D') / d = (D/d)

λ' = (D/d) * d / D

= [(0.006/2) * 10^10 / 4500] * (4500 / 0.006) Å

= 0.0045 m * 10^10 / 2 Å

= \(0.00225 * 10^{10\) Å

=\(2.25 * 10^7\)Å

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Option D: the link between the total heat flow and the temperature difference between the sample's Centre and the water at a specific time.

\(Q\:=\:m_sc_s\left(T_{s,i}-T_s\right)\)

\(T_s\right =(T_{s,i}-T_s\right))\)

\(Q=m_wc_w\left(T_w-T_{w,i}\right)\)

\(Q=m_wc_w\left(T_w-T_{w,i}\right)\)

\(T_{diff} =(T_{s}-T_w\right))\)

        = \(T_{s,i} -\frac{Q}{m_{s}C_{s}} -(T_{w,i}\right +\frac{Q}{m_{s}C_{s}} )\)

        =\((T_{s,i} - T_{w,i} )-Q(\frac{1}{m_{s}C_{s}} +\frac{1}{m_{w}C_{w}})\)

Specific time refers to a precise moment in time, often denoted by a particular time and date. It can be expressed in different ways depending on the context, such as using a 24-hour clock or the AM/PM system. Specific time is essential for scheduling events, meetings, and appointments, and for coordinating activities across different time zones. It is also crucial for time-sensitive activities such as transportation, where schedules must be coordinated down to the minute. The concept of specific time is used in many fields, including science, technology, business, and everyday life. In modern times, technologies such as smartphones and computers have made it easier than ever to track and coordinate specific times across the globe.

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The complete question is:

1. we know that the total amount of heat that flows out of the sample and into the water at a specific time is given by LaTeX: \(Q\:=\:m_sc_s\left(T_{s,i}-T_s\right)Q=mscs(Ts,i−Ts)\), where LaTeX: \(T_sTs\) is the temperature of the sample at a specific time and, again, LaTeX: \(T_{s,i}Ts,i\)is the initial temperature of the sample (at time 0). To simplify the math, we may neglect the heat leak term here to say that this is roughly the same amount of heat the flows into the water, so LaTeX: \(Q=m_wc_w\left(T_w-T_{w,i}\right)Q=mwcw(Tw−Tw,i)\), where LaTeX:\(T_wTw\) is the temperature of the water at this same specific time and LaTeX:  is the initial temperature of the water.

In the lab, we will measure both the sample and water temperatures as a function of time, but the important quantity is the difference between these temperatures since this is what drives the heat flow between the center of the sample and the water. Using the above equations (solving for the temperatures of the sample and the water bath at a particular time), we can find the relationship between the total amount of heat flow and the difference in the temperatures of the center of the sample and water at some moment in time. This yields _________________________________.

sample and water at some moment in time. This yields _________________________________.

Group of answer choices

A. \(T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)-\left(\frac{1}{m_sc_s}-\frac{1}{m_wc_w}\right)Q\)

B \(T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)+\left(\frac{1}{m_sc_s}+\frac{1}{m_wc_w}\right)Q\)

C.\(T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)+\left(\frac{1}{m_sc_s}-\frac{1}{m_wc_w}\right)Q\)

D. \(T_{dif}=T_{s\:}-T_w=\left(T_{s,i}-T_{w,i}\right)-\left(\frac{1}{m_sc_s}+\frac{1}{m_wc_w}\right)Q\)

The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

Based on the diagram, the correct statement is: Government and consumers make economic decisions in a mixed economy, while consumers make economic decisions in a market economy.

In a mixed economy, economic decisions are made by both the government and consumers.

The government plays a significant role in regulating and influencing economic activities through policies, regulations, and interventions.

In market economy, economic decisions are primarily made by consumers. The market forces of supply and demand dictate the allocation of resources, production levels, and pricing.

The freedom to buy and sell whatever they choose is what ultimately determines how commodities and services are produced and distributed.

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Answer:

The ratio is  \(\frac{F_1}{F_2} =1\)

Explanation:

From the question we are told that

    The mass of planet 1 is \(m_1\)

    The first gravitational force is \(F_1\)

     The mass of planet two is \(m_2 = 2 m_1\)

       The distance of planet 1 from the star is \(d_1\)

     The distance of planet 2 from the star is \(d_2 = 2 d_1\)

      The second planet  gravitational force is \(F_2\)

      The mass of the sun is  \(m_s\)

Generally the  gravitational force for first planet is mathematically represented as

        \(F_1 = \frac{Gm_s m_2 }{d_1^2}\)

The  gravitational force for second planet is mathematically represented as  

       \(F_2 = \frac{Gm_s m_2}{d_2^2}\)

           \(F_2 = \frac{Gm_s 2(m_1)}{ 2d_1^2}\)

           \(F_2 = \frac{2Gm_s m_1}{ 2d_1^2}\)

So     \(\frac{F_1}{F_2} = \frac{ \frac{Gm_s m_2 }{d_1^2}}{ \frac{2Gm_s m_1}{ 2d_1^2}}\)

         \(\frac{F_1}{F_2} =1\)

Since the car is accelerating at a constant rate, the distance it travels during each second of its motion will be directly proportional to the time it has been accelerating.

In the first second, the car moved a distance of 2 meters, and in the second second, it will move twice the distance of the first second, so the car will move additional distance of 2*2 = 4 meters during the second second of its motion.

The distance traveled during the second second of its motion is 1/2 * 2 = 1 meters.

A car that accelerates at a constant rate will move a distance equal to the initial velocity multiplied by time plus 1/2 the acceleration multiplied by the square of time. Since the car starts from rest, the initial velocity is zero.

Therefore, the distance traveled during the second second is 1/2 * acceleration \(* (time)^2 = 1/2 * a * t^2 = 1/2 * a * 1^2 = 1/2 * a\) Since the car moved 2.0 meters in the first second, it means the acceleration is\(2m/s^2\), and the distance traveled during the second second is 1/2 * 2 = 1 meters.

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ANSWER:

a. Velocity decreased

STEP-BY-STEP EXPLANATION:

If the person inside the bus experiences a forward movement, this means that the bus is braking, since by action-reaction to compensate for the movement, the body moves forward.

When braking, it means that there is a decrease in speed, therefore, the correct answer is a. Velocity decreased

The acceleration of the car, when it is made to stop is -13.33 m/s². The correct option is B

Acceleration is given by the ratio of resultant or total force acting on any object and the its mass.

It can also be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

A car came to a stop from a speed of 28 m/s in a time of 2.1 seconds.

The final velocity will be zero.

So, the acceleration is

a = 0-28/2.1

a = -13.33 m/s²

Thus, the acceleration of the car is -13.33 m/s².

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The linear speed will be the insect be 0.5759 meter/second carried collision with the near stationary photograph.

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

Given that an insect lands 0.1m from the center of the turn table.

Rotational speed of the turn table = 55 rev/min

= (55×2π/60) rad/second

= 5.759 rad/second.

Hence, the speed of the insect be = Rotational speed × length

= 5.759 rad/second × 0.1 M.

= 0.5759 meter/second.

Therefore, the speed of the insect be 0.5759 meter/second.

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As positively charged sodium ions enter the axon, potassium ions flow out to repolarize part of the axon.

At the beginning of an action potential of cell membrane, the sodium ion gates open and sodium ions flows into the cell.  This process is called depolarization. Due to rapid influx of sodium ion, the channel is eventually closed.

The potassium channels are then activated in a process called repolarization. This process occurs when the potassium channels open and allow potassium ions to flow out of the cell.

To maintain the cell membrane potential, cells are kept at low concentration of sodium ions and high concentration of potassium ions.

Thus, as positively charged sodium ions enter the axon, potassium ions flow out to repolarize part of the axon.

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The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

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Answer:

Every 11 years The solar cycle is the cycle that the Sun's magnetic field goes through approximately every 11 years.

Explanation:

Our Sun is a Large ball of electrically-charged hot gas. This charged gas moves, generating a powerful magnetic field. The Sun's magnetic field goes through a cycle, called the solar cycle.

Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun’s north and south poles to flip back again.

The solar cycle affects activity on the surface of the Sun, such as sunspots which are caused by the Sun's magnetic fields. As the magnetic fields change, so does the amount of activity on the Sun's surface.

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Brain-List?

Answer:

Equal reaction from the pair in every action there's an equal and opposite reaction

The item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

\(\rm F = \dfrac{Gm_1m_2}{r^2}\)

Where G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

It is given that:

An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

As we know,

An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg and the item will keep moving at a consistent speed.

Thus, the item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

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Answer:

Below in the picture:-

I hope it helps.

The power used, given that a force of 20 N is used to push the shopping cart 3.5 m in 2 seconds is 35 W

First, we shall determine the work done in pushing the cart. Details below:

Wd = Fd

Wd = 20 × 3.5

Wd = 70 J

Finally, we shall determine the power used in pushing the cart. Details below:

P = Wd / t

P = 70 / 2

P = 35 W

Thus, we can conclude that the power used is 35 W

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Answer:

20 kg

Explanation:

The kinetic energy (KE) of an object is given by:

KE = (1/2) * m * v^2

Where m is the mass of the object, and v is its velocity.

We can rearrange this formula to solve for the mass:

m = 2 * KE / v^2

Plugging in the values given:

m = 2 * 1000 J / (10 m/s)^2

m = 20 kg

Therefore, the mass of the Puffin flying at 10 m/s with 1000 J of KE is 20 kg.