Simple machines make work easier. You must left a move this load of Saul for your parents. It is heavy you’re using to simple machines when you use this wheel barrow wight simple machine included in the wheelbarrow makes it easier to move the song from one place to another.

Answer:

0.002kg and 0.01cm

Explanation:

500 leaves has a thickness is 5cm

Means I leaf has a thickness of 5/500= 0.01cm

Similarly the mass of one leaf would be 1/500 =0.002kg

Answer:

Explanation:

From the information given;

mass of the crate m = 41 kg

constant horizontal force = 135 N

where;

\(s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m\)

coefficient of kinetic friction \(u_k\) = 0.28

a)

To start with the work done by the applied force \((W_f)\)

\(W_F = F\times (s_1 +s_2) \times cos(0) \ J\)

\(W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J\)

Work done by friction:

\(W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J\)

Work done  by gravity:

\(W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j\)

Work done by normal force;

\(W_n = N \times (s_1 + s_2) \times cos (90) \ J\)

\(W_n = 0 \ J\)

b)

total work by all forces:

\(W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12\)

W = 2100.5  J

c) By applying the work-energy theorem;

total work done = ΔK.E

\(W = \dfrac{1}{2}\times m \times (v^2 - u^2)\)

\(2100.5 = 0.5 \times 41 \times v^2\)

\(v^2 = \dfrac{2100.5}{ 0.5 \times 41 }\)

\(v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}\)

ANSWER

2296.2 m

EXPLANATION

The average velocity is the quotient between the displacement and time:

Solving for displacement:

We know that they walk at an average velocity of 0.89 m/s and it took them 43 minutes to get to the store. First we have to either convert the time from minutes to seconds or convert the velocity from m/s to m per minute. Let's convert the time, because it is a bit faster:

Now we can multiply the velocity and time, because their units are consistent:

Their displacement to the store was 2296.2 meters

The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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Answer:

The SI unit of power is the watt

Explanation:

The company's predetermined overhead rates in the Molding Department and the Fabrication Department are $4.00 and $10.00 per machine-hour, respectively.

What is overhead?

Overhead refers to the indirect costs that are associated with operating a business but are not directly related to producing goods or services. These expenses include items such as rent, utilities, insurance, salaries of non-production personnel, and other costs associated with running a business. Overhead costs are typically considered fixed costs because they do not vary with the level of production or sales.

To calculate the predetermined overhead rates, we need to use the following formulas:

Molding Department predetermined overhead rate = Estimated total manufacturing overhead cost for Molding Department ÷ Estimated total machine-hours for Molding Department

Fabrication Department predetermined overhead rate = Estimated total manufacturing overhead cost for Fabrication Department ÷ Estimated total machine-hours for Fabrication Department

Using the given information, we can calculate the predetermined overhead rates as follows:

Molding Department predetermined overhead rate = $10,000 ÷ 2,500 machine-hours

= $4.00 per machine-hour

Fabrication Department predetermined overhead rate = $15,000 ÷ 1,500 machine-hours

= $10.00 per machine-hour

Therefore, the company's predetermined overhead rates in the Molding Department and the Fabrication Department are $4.00 and $10.00 per machine-hour, respectively.

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Answer:

This passage is part of the resolution because it shows what happens after the climax. It wraps up the conflict, and then the story is over.

Explanation: cause I am smart thats why

Answer:

1.8 ms to the left

Explanation:

Answer:1.8 ms to the left

Explanation:

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

Answer:

1. 125N

2. 200N

3. 280N

Explanation:

Force= mass × acceleration

1. force= 25×5

= 125N

2. Force= 50×4

=200N

3. Force=70×4

=280N

equal in magnitude but opposite in direction

Answer:

192.6N

Explanation:

Let's consider the forces acting on the beam:

Weight of the beam (W): It acts vertically downward and has a magnitude of W = mass * gravitational acceleration = 39.4 kg * 9.8 m/s^2.

Force exerted by the link on the beam (F_link): It acts at an angle of 90 degrees with respect to the beam and has two components: the vertical component and the horizontal component.

Tension in the cable (T): It supports the far end of the beam and acts at an angle of 90 degrees with respect to the beam. Since the angle between the beam and the cable is 90 degrees, the tension in the cable only has a vertical component.

Let's break down the forces acting on the beam:

Vertical forces:

W (weight of the beam) - T (vertical component of tension) = 0

T = W

Horizontal forces:

F_link (horizontal component of the force exerted by the link) = ?

To find the magnitude of the horizontal component of the force exerted by the link on the beam (F_link), we need to consider the equilibrium of forces in the horizontal direction.

Since the beam is inclined at an angle of θ = 33.1 degrees with respect to the horizontal, the horizontal equilibrium equation can be written as:

F_link = W * sin(θ)

Let's substitute the given values:

W = 39.4 kg * 9.8 m/s^2

θ = 33.1 degrees

F_link ≈ (39.4 kg * 9.8 m/s^2) * sin(33.1 degrees)

Using a calculator, we find that the magnitude of the horizontal component of the force exerted by the link on the beam (F_link) is approximately 192.6 N.

The net electric force on charge q2 is 28.7 N.

The net electric force on charge q2 is calculated by applying Coulomb's law of electrostatic force.

F(net) = F(12) + F(23)

The force on q2 due to charge 1 is calculated as;

F(12) = -(9 x 10⁹ x 8 x 10⁻⁶ x 3.5 x 10⁻⁶ )/(0.1²)

F(12) = 25.2 N

The force on q2 due to charge 3 is calculated as;

F(23) = (9 x 10⁹ x 2.5 x 10⁻⁶ x 3.5 x 10⁻⁶ )/(0.15²)

F(23) = 3.5 N

The net force on q2 is calculated as;

F(net) = 25.2 N + 3.5 N = 28.7 N

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a) A = 8.0 m South

Since the vector is directly along the South direction, there is no x component.

x component: 0 m

y component: -8.0 m (negative because it's southward)

Component form: A = (0, -8.0)

b) B = -15.0 m at 30° East of North

To find the components, we can use the following relationships:

x component: B_x = B * sin(θ)

y component: B_y = B * cos(θ)

B_x = -15.0 * sin(30°) = -15.0 * 0.5 = -7.5 m

B_y = -15.0 * cos(30°) = -15.0 * (sqrt(3)/2) ≈ -12.99 m

Component form: B ≈ (-7.5, -12.99)

c) C = 12.0 m at 25° South of West

x component: C_x = -C * cos(θ) (negative because it's westward)

y component: C_y = -C * sin(θ) (negative because it's southward)

C_x = -12.0 * cos(25°) ≈ -10.85 m

C_y = -12.0 * sin(25°) ≈ -5.16 m

Component form: C ≈ (-10.85, -5.16)

d) D = 10.0 m at 53° West of North

x component: D_x = -D * sin(θ) (negative because it's westward)

y component: D_y = D * cos(θ)

D_x = -10.0 * sin(53°) ≈ -8.0 m

D_y = 10.0 * cos(53°) ≈ 6.0 m

Component form: D ≈ (-8.0, 6.0)

The right sphere will acquire an equal and opposite net positive charge to balance the negative charge on the left sphere.

Since the left sphere is attracted to the positively charged rod, it means that the left sphere acquires a temporary negative charge due to induction.

The positive charge on the rod repels electrons in the left sphere, causing them to move away from the rod side and accumulate on the opposite side, resulting in a net negative charge on the left sphere.

According to the principle of charge conservation, the net charge on the system must remain zero. Therefore, the right sphere acquires an equal and opposite net positive charge to balance the negative charge on the left sphere.

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It will accelerate continuously. The acceleration put on by gravity will be that. Thus, correct option is: It has the same acceleration as it moves up and down the plane.

The force that pulls items toward the centre of a planetary or other body is known as gravity. The gravitational pull maintains each of the planet in orbit about the sun. The degree to which a coordinate in space-time is invisibly bent determines the intensity of the gravitational "field" at any given place in space or time. Massive items tumble down these bends in the direction of one another. The stronger the attraction between items, the more matter there is, and the closer together things are. And unlike electricity and magnetism, which may either repel or attract, gravity always draws objects together.

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The complete question is as follows:

The mass of the ball is approximately 10.24 kg, found using the conservation of energy principle.

What is the mass of a ball thrown upward with a speed of 10 m/s from a height of 20 m, using the conservation of energy principle?

To find the answer, we can use the conservation of energy principle, which states that the total energy of a system remains constant if there is no net work done on the system.

Initially, the ball has some potential energy due to its height above the ground and kinetic energy due to its motion. As it travels upwards, its potential energy increases while its kinetic energy decreases until it reaches its highest point, at which its kinetic energy is zero and its potential energy is at its maximum. Then, as it falls back down, its potential energy decreases while its kinetic energy increases until it reaches the ground, at which its potential energy is zero and its kinetic energy is at its maximum.

Let's use the following variables:

m = mass of the ball

v_i = initial velocity of the ball (10 m/s)

v_f = final velocity of the ball (0 m/s at the highest point)

h = height of the ball above the ground (20 m)

g =  (9.81 m/s^2)

The total energy of the ball initially is:

E_i = 1/2 * m * v_i^2 + m * g * h

At the highest point, the total energy of the ball is:

E_f = 1/2 * m * v_f^2 + m * g * 2h

Since the ball is at rest at the highest point, its final velocity is zero, so:

E_f = m * g * 2h

By the conservation of energy principle, the total energy of the ball is constant, so:

E_i = E_f

Substituting the expressions for E_i and E_f and solving for the mass of the ball:

1/2 * m * v_i^2 + m * g * h = m * g * 2h

1/2 * v_i^2 + g * h = 2g * h

m = (1/2 * v_i^2 + g * h) / g

m = (1/2 * 10^2 + 9.81 * 20) / 9.81

m = 10.24 kg

Therefore, the mass of the ball is approximately 10.24 kg.

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In a collision that is completely inelastic, the two objects cling together and continue to move as a single entity. As the two objects' ultimate velocities are equal, v′1=v′2=v′ (v 1 ′ = v 2/ ′ = v),

Let's move on to the second category of collisions. Kinetic energy is not maintained in an inelastic collision. When two things collide in a collision that is precisely inelastic, the max amount of mechanical energy is lost as the objects cling together again after impact.

When some of the angular momentum of a colliding item or system is lost, a physics term known as an elastic collisions is used. In a completely inelastic collision, the interacting atoms stay together and the most kinetic energy is wasted. Such situations involve the use of wasted kinetic energy to bind the two items together.

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The higher the dissociation constant the stronger the acid or base. Since electrolytes are created as ions are freed into solution there is a relationship between the strength of an acid, a base, and the electrolyte it produces. Acids and bases are measured using the pH scale.

Answer: Should the sabre tooth curriculum really exist at the moment? As far as I am concerned, the sabre tooth program should remain, because our school system needs to help people know how to do complex activities on their own and we let them do activities that will help them develop their brains as well as their skills.

Great schools promote better teaching because they have the most experience of teaching practice and are accountable for executing the instruction in the classroom. Teacher engagement is also vital for the effective and substantive implementation of the curriculum.

Sources: files.eric.ed.gov and James Kennedy Monash

We have that the Force  is mathematically given as

F=170.833N

Question Parameters:

Generally the equation for the Force   is mathematically given as

\(F=\frac{T}{d}\)

Therefore

F=41/0.24

F=170.833N

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Answer: The scenario violates the second law of thermodynamics.

Explanation: The second law states that heat cannot be converted into work without some loss of usable energy, and that the amount of usable energy in a closed system will always decrease over time. Therefore, the machine described in the scenario cannot exist because it would violate the second law by converting all of the heat into electricity without any loss of usable energy.

Answer:

Radius of curvature of the mirror = 250 cm

Explanation:

Given:

Object distance from mirror = 250 cm (u=-250)

Object distance appears in mirror = 250 cm (v=-250)

Find:

Radius of curvature of the mirror

Computation:

Using mirror formula

1/f = 1/v + 1/u

1/f = 1/(-250) + 1/(-250)

f = (-250/2)

f = -125 cm or 125 cm

Radius of curvature of the mirror = 2(f)

Radius of curvature of the mirror = 2(125)

Radius of curvature of the mirror = 250 cm

Answer:

All of the objects in the solar system orbit the

Explanation:

sun

It takes approximately 9.05 seconds to roll the drum a distance of 36 meters.

We can use the formula for the circumference of a circle:

Circumference = 2 * π * radius

Given:

Radius (r) = 0.40 m

Circumference (C) = 2 * π * 0.40 m

We must figure out how many full rotations the drum makes to go 36 meters in order to calculate how long it takes to roll the drum. Since we are aware of the circumference, we can determine the number of full turns as follows:

Number of turns = Distance / Circumference

Given:

Distance = 36 m

Number of turns = 36 m / (2 * π * 0.40 m)

Now that we know how many turns there are, we can calculate the time by multiplying that number by the length of a turn, which is given as 8.0 seconds:

Time = Number of turns * Time per turn

Time = (36 m / (2 * π * 0.40 m)) * 8.0 s

By substituting the values into the equation, we can calculate the time:

Time = (36 / (2 * 3.14159 * 0.40)) * 8.0 s

Time ≈ 9.05 s

So, it takes approximately 9.05 seconds to roll the drum a distance of 36 meters.

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Answer:

B. 1990

This article is part of the NASA Knows (Grades 5-8) series. The Hubble Space Telescope is a large telescope in space. It was launched into orbit by space shuttle Discovery on April 24, 1990. Hubble orbits about 547 kilometers (340 miles) above Earth.

Answer: 1990

Explanation: Wikipedia say “The Hubble Space Telescope (often referred to as HST or Hubble) is a space telescope that was launched into low Earth orbit in 1990 and remains in operation. It was not the first space telescope, but it is one of the largest and most versatile, renowned both as a vital research tool and as a public relations boon for astronomy. The Hubble telescope is named after astronomer Edwin Hubble and is one of NASA's Great Observatories, along with the Compton Gamma Ray Observatory, the Chandra X-ray Observatory, and the Spitzer Space Telescope.”

a. traction

If your vehicle has a high center of gravity or if the surface traction is less than ideal, it is important to enter a curve slower than the posted speed. This is because vehicles with a high center of gravity or poor surface traction are more prone to instability and are more likely to roll over or lose control when negotiating sharp turns or curves at high speeds.

Traction refers to the friction or grip between the tires of a vehicle and the road surface. Poor surface traction can be caused by a variety of factors, such as wet or slippery roads, loose gravel or debris, or worn or damaged tires. When surface traction is poor, it becomes more difficult for the tires to maintain contact with the road and to transmit the forces of acceleration, braking, and steering to the road.

By entering a curve slower than the posted speed, you can reduce the lateral forces acting on your vehicle and give your tires more time to grip the road. This can help you maintain control and stability, and reduce the risk of accidents or injuries. It is important to always adjust your speed and driving style to suit the conditions of the road and the capabilities of your vehicle.

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A transformer in which the output (secondary) voltage is greater than its input (primary) voltage is called a step-up transformer.

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Organelles are specialized structures within cells that perform specific functions, such as energy production, protein synthesis, and waste removal.

Some examples of organelles include mitochondria, which produce energy for the cell, and ribosomes, which are involved in protein synthesis.

The size of cells can vary widely depending on the organism and the type of cell. For example, human cells can range from 10 to 30 micrometers in diameter, while bacterial cells are typically much smaller, ranging from 1 to 5 micrometers in diameter.

In summary, organelles are specialized structures within cells that perform specific functions, and the size of cells can vary widely depending on the organism and the type of cell.

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Answer:

The disk covers a rotation of 90º after 3 seconds.

Explanation:

Since the disk rotates at constant angular speed, we can determine the change in angular position (\(\Delta \theta\)), measured in sexagesimal degrees, by the following kinematic formula:

\(\Delta \theta = \omega\cdot \Delta t\) (1)

Where:

\(\omega\) - Angular velocity, measured in sexagesimal degrees per second.

\(\Delta t\) - Time, measured in seconds.

If we know that \(\omega= 30\,\frac{\circ}{s}\) and \(\Delta t = 3\,s\), then the change in angular position is:

\(\Delta \theta = \left(30\,\frac{\circ}{s} \right)\cdot (3\,s)\)

\(\Delta \theta = 90^{\circ}\)

The disk covers a rotation of 90º after 3 seconds.