There are two objects (A and B) each with its own mass and charge. The electrostatic force between them has a value of 23.2N and is attractive. The two objects are separated by a distance of 2.3cm with object A being to the left of B. Assume one charge has 4 times the charge of the other.
a. Do the two charges have the same signs or opposite signs?
b. IF the two charges have opposite signs, is charge A or B negative?
c. Which charge (A or B) has the higher value?
d. Find the value of each charge.

The boy throws the ball horizontally. The initial velocity of the ball as it left the boy's hand was approximately 3.72 m/s.

Initial velocity, often represented as v0, is the velocity of an object at the beginning of a time interval or at the start of a motion.

Use the following kinematic equations to arrive at the answer:

Horizontal velocity (Vx) = Distance / Time

Vertical displacement (y) = V0y*t + (1/2)gt²

Vertical velocity (Vy) = V0y + g*t

Tan(theta) = Vy / Vx

where V0y is the initial vertical velocity, g is acceleration due to gravity (9.8 m/s²), and theta is the angle of inclination.

First, let's find the time it takes for the ball to hit the ground. We can use the vertical displacement equation and set y = 0:

0 = V0y*t + (1/2)gt²

Simplifying and solving for t, we get:

t = sqrt((2y) / g)

= sqrt((21.10 m) / 9.8 m/s²)

= 0.472 s

Now, we can use the horizontal velocity equation to find Vx. Since the ball was thrown horizontally, Vx is the same as the initial velocity (V0):

Vx = Distance / Time

= (horizontal distance travelled by ball) / t

We don't know the horizontal distance travelled by the ball, but we can find it using the vertical displacement equation. At the instant the ball hits the ground, its vertical displacement (y) is:

y = V0y*t + (1/2)gt²

= 0 + (1/2)gt²

= (1/2)*9.8 m/s² * (0.472 s)²

= 1.10 m

This means the ball travelled a total distance of:

distance = horizontal distance + vertical distance

= x + 1.10 m

where x is the horizontal distance travelled by the ball. We can find x using the angle of inclination and the vertical displacement:

Tan(theta) = Vy / Vx

Vy = V0y + g*t

Solving for V0y, we get:

V0y = Vy - g*t

Plugging in the numbers, we get:

V0y = Tan(theta) * Vx - g*t

= Tan(30 deg) * Vx - 9.8 m/s² * 0.472 s

= 0.577 * Vx - 4.62 m/s

Now, we can use the vertical displacement equation again to find x:

y = V0yt + (1/2)gt²

= (0.577Vx - 4.62 m/s) * 0.472 s + (1/2)*9.8 m/s² * (0.472 s)²

= 1.10 m

Simplifying and solving for Vx, we get:

Vx = (2y - 0.577V0t) / t

= (21.10 m - 0.577*(0.577*Vx - 4.62 m/s)*0.472 s) / 0.472 s

= 3.72 m/s

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The truck's acceleration is 3.0m/s² and the momentum of the truck is  144000 kg m/s.

It is the rate at which the speed and direction of a moving object vary over time.

We can use the following equation to calculate the acceleration of the truck:

a = (v - u) / t

where

a = acceleration

v = final velocity = 18.0 m/s

u = initial velocity = 0 m/s (the truck starts from rest)

t = time taken = 6.00 s

Substituting the values, we get:

a = (18.0 m/s - 0 m/s) / 6.00 s

a = 3.00 m/s²

Therefore, the acceleration of the truck is 3.00 m/s².

We can use the following equation to calculate the momentum of the truck:

p = m * v

where

p = momentum

m = mass of the truck = 8000.0 kg

v = final velocity = 18.0 m/s

Substituting the values, we get:

p = 8000.0 kg * 18.0 m/s

p = 144000 kg m/s

Therefore, the momentum of the truck after it has reached its final velocity is 144000 kg m/s.

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Answer:

Explanation:

We can solve this problem using kinematic equations. We know that the initial velocity of the ball is 55 m/s at an angle of 30° to the horizontal. We can break this velocity into its horizontal and vertical components:

vx = v0 cos θ = 55 cos 30° = 47.6 m/s

vy = v0 sin θ = 55 sin 30° = 27.5 m/s

We can now use the vertical motion equation to find the time it takes for the ball to reach its maximum height:

Δy = vy t + 0.5 a t^2

At the maximum height, the vertical velocity of the ball is 0, so we have:

0 = vy + a t_max

Solving for t_max, we get:

t_max = -vy / a = -27.5 / (-9.8) = 2.81 s

The ball will take twice this time to reach the fence, since it needs to come back down to the ground:

t_total = 2 t_max = 5.62 s

The horizontal distance the ball travels during this time is:

Δx = vx t_total = 47.6 × 5.62 = 267.7 m

Since this distance is greater than the distance to the fence (120 m), the ball will reach the fence if it leaves the bat traveling upward at an angle of 30° to the horizontal.

Answer:

Explanation:

To choose a material for the handrails of the playhouse, we need to consider its strength and durability. One option could be stainless steel, which has a high tensile strength and is resistant to corrosion and weathering. Another option could be treated wood, which is also strong and can be treated to resist moisture and insects. Ultimately, the choice would depend on factors such as cost, aesthetics, and availability of materials.

that the relative placements of the charges as well as their multiples affect a set of ions' potential energy. When the specific charge have the same sign or have equal signs, the energy is positive. Or else, it is negative.

The force acting just on two objects affects the potential energy formula. The formula for gravitational force is P.E. (= mgh, where g seems to be the acceleration caused by gravity (9.8 m/s2 at the earth's surface) while h represents the elevation in metres.

As a result, the system's potential energy equals the sum of a work that was done to set up the entire system of two counts. The potential energy that exists in the combination of two charges in such an external field can be stated as follows: q1V(r1) = q2V(r2) + (q1q2/4or12).

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The tension in the rope is 196.2 N. The rope is exerting a force of 196.2 N on the object to keep it stationary.

Assuming that the mass is not accelerating, the tension in the rope must be equal to the weight of the mass. The weight of the mass can be found using the formula:

weight = mass x acceleration due to gravity

where acceleration due to gravity is approximately 9.81 m/s².

Therefore, the weight of the mass is:

weight = 20 kg x 9.81 m/s² = 196.2 N

Since the mass is held stationary, the tension in the rope must be equal to the weight of the mass, which is 196.2 N. So the tension T in the rope is 196.2 N.

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Answer:

In this problem, the harp string is fixed at both ends, so it is a standing wave with nodes at both ends and an antinode in the center. The frequency of the wave is given as 440 Hz, and the length of the string is 30.5 cm.

(a) To find the wavelength of the wave, we can use the formula:

λ = 2L/n

where λ is the wavelength, L is the length of the string, and n is the number of nodes. In this case, n = 2 (since there are nodes at both ends) and L = 30.5 cm, so we have:

λ = 2(30.5 cm)/2 = 30.5 cm

Therefore, the wavelength of the wave is 30.5 cm.

(b) To find the speed of the wave on the string, we can use the formula:

v = fλ

where v is the speed of the wave, f is the frequency of the wave, and λ is the wavelength. In this case, f = 440 Hz and λ = 30.5 cm, so we have:

v = (440 Hz)(30.5 cm) = 13420 cm/s

Therefore, the speed of the wave on the string is 13420 cm/s

Explanation:

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As a result, the combined Honda-Toyota system's post-collision speed is 7.5 m/s.

The system's center of mass was v/2 before to the collision since one automobile had a velocity of v and the other zero. The total momentum is equal to the entire mass times the velocity of the center of mass, or (2m)(v/2) = mv before and after.

Initial momentum of Honda = m1 * v1

= 200 kg * 30 m/s

= 6000 kg·m/s

Final momentum of combined system = (m1 + m2) * v_final

Setting the two momenta equal to each other, we get:

6000 kg·m/s = 800 kg * v_final

Solving for v_final, we get:

v_final = 6000 kg·m/s / 800 kg

= 7.5 m/s

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There are two forces acting on the rock: the force of gravity pulling it downward and the force of the boulder supporting it from underneath.

The force of gravity is the gravitational attraction between the rock and the Earth. It pulls the rock downward with a force equal to its weight, which is given by the equation Fg = mg, where Fg is the force of gravity, m is the mass of the rock, and g is the acceleration due to gravity (approximately 9.81 m/s^2).

The concept of forces explains why the rock stays on top of the boulder because the forces are balanced. The force of gravity pulling the rock downward is equal and opposite to the force of the boulder supporting it from underneath. As a result, the rock remains in equilibrium, or a state of balance, on top of the boulder. If either force were to change, the equilibrium would be disrupted, and the rock would either fall to the ground or be pushed off the boulder.

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Answer:

Explanation:

a. The compact car going 70 MPH has more kinetic energy than the tractor trailer going 15 MPH. Kinetic energy is proportional to the square of the velocity, so even though the tractor trailer may have more mass, the higher velocity of the compact car results in greater kinetic energy.

b. The SUV and the pickup truck have the same kinetic energy since they have the same mass and velocity.

c. The compact car going 45 MPH has the most kinetic energy since it has the highest velocity out of the three vehicles. The SUV going 35 MPH has less kinetic energy, and the school bus going 15 MPH has the least kinetic energy.

Answer:

Explanation:

What is the gravitational force between them?

To calculate the gravitational force between two objects, we can use the formula:

F = G * (m1 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (6.6743 x 10^-11 N * m^2 / kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.

Plugging in the given values, we get:

F = (6.6743 x 10^-11 N * m^2 / kg^2) * (2000000 kg) * (3000000 kg) / (50 m)^2

F = 0.8046 N

Therefore, the gravitational force between the two asteroids is approximately 0.8046 N.

The light waves reflecting off a red stop sign would be longer than the light waves reflecting off a violet-colored jacket. This is because red light has a longer wavelength than violet light.

Light waves, like all waves, are characterized by their wavelength. The wavelength of a wave determines its color, with shorter wavelengths appearing as blue and violet, and longer wavelengths appearing as red and orange.

Because the wavelength of red light is longer than the wavelength of violet light, the light waves reflecting off the red stop sign would be longer than the light waves reflecting off the violet-colored jacket.

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If F₁ has a greater magnitude than F₂, the box will accelerate backward because the net force is in the backward direction (1st option)

To obtain the direction in which the box will move, we shall determine the net force acting on the box. This is illustrated below:

Assumption:

Net force = Magnitude of force 1 (F₁) - Magnitude of force 2 (F₂)

Net force = F₁ - F₂

Net force = 40 - 25

Net force = 15 N backward

From the above illustration, we can see that the net force is 15 N backward.

Thus, we can conclude from the box will accelerate backward (1st option)

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Answer: D. It needs a medium to travel.

Explanation:

One way to categorize waves is on the basis of the direction of movement of the individual particles of the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three notable categories: transverse waves, longitudinal waves, and surface waves.

They are specifically made tο be ideal fοr cοnducting live electrical lines because οf their electrical insulatiοn and mechanical tοughness. A structure called an electric pylοn οf hοt-rοlled steel bevels οr gusset plates.

Other materials, such as cοncrete and wοοd, may alsο be utilised in additiοn tο steel. Transmissiοn tοwers can be divided intο fοur main categοries: suspensiοn, terminal, tensiοn, οr transpοsitiοn.

This Central Electricity Bοard held a cοmpetitiοn in 1927, and the winning entry was chοsen by the classical designer Sir Reginald Blοοmfield. He settled οn an A-frame structure with latticewοrk that was οffered by the American cοmpany Milliken Brοthers and is still in use tοday.

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Complete question:

Answer: 4 N backwards

Explanation:

The work done by the gaseous system if the volume is increased to 0.12 cubic meters is given as 21,600 joules

To find the work done by the gas, we can use the formula:

W = PΔV

where W is the work done, P is the pressure of the gas, and ΔV is the change in volume.

At the initial state, the pressure is P = 2.7 × 10^5 Pa and the volume is V1 = 0.04 m^3. At the final state, the volume is V2 = 0.12 m^3.

The change in volume is ΔV = V2 - V1 = 0.12 m^3 - 0.04 m^3 = 0.08 m^3.

Substituting these values into the formula, we get:

W = PΔV = (2.7 × 10^5 Pa) × (0.08 m^3) = 21,600 J

Therefore, the work done by the gaseous system is 21,600 joules (J).

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Answer:

4.6x10^4 joules

Explanation:

Answer:

The speed of the ball just before it hits the crossbar is 7.22 m/s.

Explanation:

We can use the conservation of energy to solve this problem.

At the moment the player kicks the ball, the ball has only kinetic energy, since it was at rest on the ground before being kicked. When the ball hits the crossbar, it has both kinetic energy and potential energy, since it is at a height above the ground. We can set the initial kinetic energy equal to the sum of the final kinetic and potential energy:

(1/2)mv^2 = mgh

where:

m = mass of the ball (0.500 kg)

v = initial speed of the ball (15.0 m/s)

g = acceleration due to gravity (9.80 m/s^2)

h = height of the crossbar above the ground (2.6 m)

We want to solve for the speed of the ball just before it hits the crossbar, which we can do by rearranging the equation:

v = sqrt(2gh)

v = sqrt(29.802.6) = 7.22 m/s (rounded to two decimal places

newton 3th law of motion and newton's law of universal gravitation

Answer: 1. Newton's First Law of Motion (Law of Inertia): An object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.

During liftoff, the spacecraft is initially at rest. However, the rocket engines generate a force that propels the spacecraft forward and overcomes its initial state of rest. Once the spacecraft is in motion, it will continue to move forward with a constant velocity unless acted upon by other external forces, such as air resistance or gravity.

2. Newton's Second Law of Motion: The acceleration of an object is directly proportional to the force applied to it, and inversely proportional to its mass.

As the rocket engines burn fuel, they generate a force that propels the spacecraft forward. The acceleration of the spacecraft is directly proportional to the force generated by the engines, and inversely proportional to the mass of the spacecraft. As fuel is consumed and the spacecraft becomes lighter, its acceleration will increase, allowing it to reach escape velocity and enter space.

3. Newton's Third Law of Motion: For every action, there is an equal and opposite reaction.

During liftoff, the rocket engines generate a powerful force that propels the spacecraft forward. However, the engines also generate an equal and opposite reaction force, pushing back against the rocket and causing it to shake and vibrate. This force is also responsible for the loud noise and exhaust plumes that are visible during liftoff.

These are the three laws of motion developed by Sir Isaac Newton, and they explain how objects move and interact with one another. They can be observed in the launch and flight of a spacecraft, from the initial state of rest to the forces that drive it forward, to the equal and opposite forces that shake the rocket during liftoff.

The second possibility is that the firefighters can position their cannon anywhere within a range of 60.2 m from the building to hit the blaze at 10.0 m above ground level.

What is Horizontal Velocity?

Horizontal velocity is the component of velocity in the horizontal direction, perpendicular to the vertical direction. It describes the speed and direction of motion of an object in the horizontal plane. In the absence of external forces, the horizontal velocity of an object moving through the air remains constant, as there is no force acting on the object in the horizontal direction.

Let's first find the time it takes for the water to reach the maximum height:

The vertical component of the initial velocity is:

vy = v * sin(θ) = 29.0 m/s * sin(53.0°) = 22.7 m/s

Using the kinematic equation:

where Δy = 10.0 m, vyi = 22.7 m/s, and a = -9.81 m/[tex]s^{2}[/tex] (the negative sign indicates that acceleration is in the opposite direction to the initial velocity).

We get:

10.0 m = 22.7 m/s * t - 1/2 * 9.81 m/[tex]s^{2}[/tex] * [tex]t^{2}[/tex]

Simplifying and solving for t, we get:

t = 1.61 s

Now, let's find the horizontal displacement of the water:

The horizontal component of the initial velocity is:

vx = v * cos(θ) = 29.0 m/s * cos(53.0°) = 18.7 m/s

Using the equation:

Δx = vx * t

where Δx is the horizontal displacement, vx is the horizontal component of the initial velocity, and t is the time we found above.

We get:

Δx = 18.7 m/s * 1.61 s = 30.1 m

So the firefighters should position their cannon 30.1 m away from the building.

To find the second possibility, we need to find the range of the water cannon, which is the horizontal distance traveled by the water before it hits the ground. The range can be calculated using the formula:

where R is the range, v is the initial velocity, θ is the angle above the horizontal, and g is the acceleration due to gravity.

Plugging in the values we get:

R = [tex]29.0^{2}[/tex] * sin(2 * 53.0°) / (2 * 9.81) = 60.2 m

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According to Gauss' equation, the total flux of an electric field in a confined surface is directly proportional to the charge enclosed.

1)To determine the outward electric flux through the rounded "side" of the cylinder, we can use Gauss's law. We choose a cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The electric field due to the sphere is spherically symmetric, so the electric field lines are parallel to the cylinder's axis and perpendicular to its sides.

E = (1/4πϵ0) (Q/r^2)

where r is the distance from the origin (center of the sphere) to the point on the Gaussian surface.

The area element of the Gaussian surface is dA = 2πRdz, where dz is an element of length along the cylinder's axis. The electric flux through the top and bottom surfaces of the Gaussian surface is then given by:

Φ = ∫E⋅dA = E ∫dA = E(2πR)L

Substituting the expression for the electric field, we have:

Φ = (Q/2ϵ0r^2)(2πRL)

Therefore, the outward electric flux through the rounded "side" of the cylinder is:

Φ = (Q/ϵ0)(R/Lr^2)

2)To determine the electric flux upward through the circular cap at the top of the cylinder, we use a flat Gaussian surface with radius R and height r, centered at the top of the cylinder. The electric field due to the charged sphere is perpendicular to the Gaussian surface, so the electric flux through the top cap is simply the flux through the flat Gaussian surface. The electric field at any point on the Gaussian surface is given by Coulomb's law as:

E = (1/4πϵ0) (Q/R^2)

The area element of the Gaussian surface is dA = πR^2, so the electric flux through the top cap is given by:

Φ = ∫E⋅dA = E ∫dA = EπR^2

Substituting the expression for the electric field, we have:

Φ = (Q/ϵ0)(R/r^2)

3)To determine the electric flux downward through the circular cap at the bottom of the cylinder, we use a similar flat Gaussian surface with radius R and height r, centered at the bottom of the cylinder. The electric flux through the bottom cap is also given by:

Φ = (Q/ϵ0)(R/r^2)

4)Adding the results from parts 1-3, we have the total outward electric flux through the closed cylinder as:

Φ_total = Φ_side + Φ_top + Φ_bottom

= (Q/ϵ0)(R/Lr^2) + 2(Q/ϵ0)(R/r^2)

Simplifying this expression, we have:

Φ_total = (Q/ϵ0) [(2R/r^2) + (R/Lr^2)]

5)According to Gauss's law, the total outward electric flux through a closed surface is proportional to the total charge enclosed within that surface. In this case, the closed surface is the cylindrical Gaussian surface with radius r and length L, centered at the origin (where the charged sphere is located). The charge enclosed within this surface is simply the charge of the sphere, which is +Q. Therefore, we expect the total outward electric flux through the closed cylinder to be:

Φ_total = Q/

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The translational speed of the barrel at the bottom of the hill is 28.1 m/s.

Translational speed is the speed of an object in a straight line. It is different from rotational speed, which is the speed of an object’s rotation. Translational speed is a measure of how quickly an object is moving in a specific direction. It is calculated by dividing the distance traveled by the time it took to travel that distance.

The barrel's initial potential energy can be calculated using the equation U = mgh, with m being the mass of the barrel (21.0 kg),
g being the acceleration due to gravity (9.80 m/s2),
and h being the height of the barrel above the bottom of the hill (31.0 m). Therefore, the barrel's initial potential energy is U = 21.0 kg × 9.80 m/s2 × 31.0 m = 6259.8 J.
At the bottom of the hill, the barrel's potential energy is zero, since it is at the lowest point.
Therefore, the barrel's total mechanical energy is equal to its kinetic energy.
Since the kinetic energy of an object is given by K = ½mv2,
where m is the mass of the barrel and v is its velocity,
we can calculate the barrel's velocity at the bottom of the hill by rearranging the equation to v = √(2K/m).
Substituting in the values for the barrel's mass (21.0 kg) and its total mechanical energy (6259.8 J) gives us v = √(2 × 6259.8 J / 21.0 kg) = 28.1 m/s.
Therefore, the translational speed of the barrel at the bottom of the hill is 28.1 m/s.

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Answer:

Explanation:

The electric flux through a surface is given by the equation:

Φ = EAcos(θ)

where Φ is the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

We are given Φ = 4.44 N⋅m2/C, A = 6.66×10−4 m2, and θ = 60.0∘. Substituting these values into the equation above and solving for E, we get:

E = Φ / (Acos(θ))

= 4.44 N⋅m2/C / (6.66×10−4 m2cos(60.0∘))

= 1.62×10^4 N/C

Therefore, the magnitude of the electric field is 1.62×10^4 N/C.

The magnitude of the electric field is 13,320 N/C.

The electric flux through a surface is defined as the product of the electric field and the area of the surface projected perpendicular to the electric field. Mathematically, we can write:

Φ = EAcos(θ)

where Φ is the electric flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

Here in the Question,

We are given the electric flux Φ = 4.44 N·m^2/C, the area A = 6.66×10^-4 m^2, and the angle θ = 60.0°. We can solve for the magnitude of the electric field E by rearranging the equation as follows:

E = Φ / (A*cos(θ))

Substituting the given values, we get:

E = 4.44 N·m^2/C / (6.66×10^-4 m^2*cos(60.0°))

Simplifying the denominator, we get:

E = 4.44 N·m^2/C / (6.66×10^-4 m^2*0.5)

E = 13,320 N/C

Therefore, 13,320 N/C is the magnitude of the electric field.

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When the box is at rest, the tension in the rope is equal to the weight of the box. The tension in the rope is 588 N.

In physics, tension refers to the pulling force that is transmitted through a string, cable, rope, etc when it is pulled tight by forces acting at both ends. Tension is a vector quantity, and measured in units of newtons (N) or pounds (lbs).

A. When the box is at rest, the tension in the rope is equal to the weight of the box, which is given by:

Tension = Weight of the box = mg = (60.0 kg)*(9.81 m/s²) = 588 N

Thus, the tension = 588 N.

B. When the box moves up at a steady 4.80 m/s, the tension in the rope is equal to the force required to lift the box against gravity, which is given by:

Tension = Weight of the box + Force to lift the box = mg + ma = (60.0 kg)*(9.81 m/s²) + (60.0 kg)*(4.80 m/s²) = 1,167.6 N

Therefore, the tension in the rope is 1,167.6 N.

C. When the box has velocity along the y-axis = 5.00 m/s and is speeding up at 5.40 m/s², the tension in the rope is given by the equation:

Tension = Weight of the box + Force to accelerate the box = mg + ma = (60.0 kg)*(9.81 m/s²) + (60.0 kg)*(5.40 m/s²) = 1,199.4 N

Therefore, the tension in the rope is 1,199.4 N.

D. When the box has velocity along y-axis = 5.00 m/s and is slowing down at 5.40 m/s², the tension in the rope is given by the equation:

Tension = Weight of the box - Force to decelerate the box = mg - ma = (60.0 kg)(9.81 m/s²) - (60.0 kg)(5.40 m/s²) = 981.6 N

Therefore, the tension in the rope is 981.6 N.

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Plugging in the values of the length and diameter, along with the error value, the error in the volume of the cylinder is approximately 1.27 cm³.

Error in measurement refers to the deviation or difference between the true or expected value and the measured value of a physical quantity. The presence of errors in measurement can affect the accuracy and precision of the results obtained.

To compute the possible error in the volume of the cylinder, we first need to calculate the volume of the cylinder using the measured values of its length and diameter:

V = πr²h

r = d/2 = 2.1/2 = 1.05 cm

V = π(1.05)²(8.9) = 31.79 cm³

Now, we need to determine the possible error in the volume, which can be calculated using the formula:

ΔV = V × √[(Δd/d)² + (Δh/h)²]

where Δd and Δh are the uncertainties in the diameter and length measurements, respectively. Substituting the given values, we get:

ΔV = 31.79 × √[(0.1/2.1)² + (0.1/8.9)²] = 1.27 cm³ (approx.)

Therefore, the possible error in the volume of the cylinder is approximately 1.27 cm³.

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Answer:

600kW

Explanation:

Power= Workdone/ time

           = 1500/8*320

           = 1500*40

           = 60000J/s

           = 600kW

    Workdone= Fd

                     = 3000*1*1/16

                     = 1500/8

                     = 750/4

                    = 137. 5Nm

3000F/320

=150F/16

s=ut+1/2at^2

3000= 1/2at^2

6000= at^2

6000/a=t^2

F=ma

20m/t=ma

20/t= a

20m=Ft

20m=F(320)

m= 8F

F=ma

= 20/tm

20m/t= 20/tm

m= 1/m

m=1kg

6000/a= 400/a^2

16= 1/a

a= 1/16ms-2

t= 20/1/16

t= 320 s

(v-u)/t=a

v= 20ms-1

Answer:

Explanation:

Since the stove is moving with a constant velocity, we know that the net force on the stove is zero. Therefore, the force of friction acting on the stove must be equal in magnitude and opposite in direction to the horizontal force being applied by the contractor. We can use Newton's second law to solve for the normal force and force of friction:

ΣF = ma

where ΣF is the net force, m is the mass of the stove, and a is the acceleration of the stove (which is zero in this case).

First, we need to convert the velocity to m/s and the forces to Newtons (N):

18 cm/s = 0.18 m/s

85 N [fwd] - force applied by contractor

447 N [down] - force of gravity on the stove

Now we can solve for the normal force:

ΣFy = 0 (since the stove is not accelerating in the y-direction)

FN - 447 N = 0

FN = 447 N

Therefore, the normal force acting on the stove is 447 N.

Next, we can solve for the force of friction:

ΣFx = 0 (since the stove is moving at a constant velocity)

Ff - 85 N = 0

Ff = 85 N [bkwd]

Therefore, the force of friction acting on the stove is 85 N [bkwd].

Finally, we can solve for the total force applied by the floor:

ΣF = ma = 0 (since the stove is not accelerating)

Ffloor - 85 N - 447 N = 0

Ffloor = 532 N [up]

Therefore, the total force applied by the floor on the stove is 532 N [up].