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Elza [17]
3 years ago
8

2. You decide to lift a 25 kg box to a height of 3 meters. How much work will you do while moving

Physics
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

W = 735.75[J]

Explanation:

Work is defined as the product of force by distance. Therefore we can use the following equation.

W=F*d

where:

W = work [J] (units of Joules)

F = force [N] (units of Newtons)

d = distance = 3 [m]

But first, we must determine the force that is equal to the product of mass by gravity (weight of the body).

F=m*g\\F=25*9.81\\F=245.25[N]

W=F*d\\W=245.25*3\\W=735.75[J]

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The disks you will be using in the lab have the following parameters. All disks are solid with outer radius R1 = 0 0632 m and in
Nesterboy [21]

Answer:

MT disc   I = 2,752 10-3 kg m²

MB disc   I = 2,726 10⁻³ kg m²

Explanation:

The moment of inertia given by the expression

        I = ∫ r² dm

for bodies with high symmetry it is tabulated

  for a hollow disk it is

        I = ½ M (R₁² + R₂²)

let's apply this equation to our case

disc MT = 1,357 kg

         I = ½ 1,357 (0.0079² + 0.0632²)

         I = 2,752 10-3 kg m²

disk MB = 1,344 kg

         I = ½ 1,344 (0.0079² + 0.0632²)

         I = 2,726 10⁻³ kg m²

4 0
3 years ago
A standing wave is established on a string that is fixed at both ends.
oksano4ka [1.4K]

Answer:

d. The length of the string is equal to one-half of a wavelength.

Explanation:

For a standing waves vibrating with Fundamental Frequency will be vibrate in one loop so  Length 2 L =  λ    ⇒  L = 1/2 λ

7 0
3 years ago
Read 2 more answers
For a science project, Janet performs four experiments that are supposed to show a chemical reaction. She displays her results i
mixas84 [53]

number 4 should be no

3 0
3 years ago
What is the advantage in solving motion problems using energy conservation principles instead of free body diagrams
riadik2000 [5.3K]

Answer:

However, the disadvantages are:

1. Many atimes for some motion prolems, free-body diagrams has to be drawn many times so to have enough equations to solve for the unknowns. This is not the same with energy conservation principles.

2. In situations where we need to find the internal forces acting on an object, we can't truly solve such problems using free-body diagram as it captures external forces. This is not the same with energy conservation principles.

Explanation:

Often times the ideal method to use in solving motion problem related questions are mostly debated.

Energy conservation principles applies to isolated systems are useful when object changes their positions in moving upward or downward converts its potential energy due to gravity for kinetic energy, or the other way round. When energy in a system or motion remains constant that is energy is neither created nor destroyed, it can therefore be easier to calculate other unknown paramters like in the motion problem velocity, distance bearing it in mind that energy can only change from one type to another.

On the other hand, free body diagram which is a visual representation of all the forces acting on an object including their directions has so many advantages in solving motion related problems which include finding relationship between force and motion in identifying the force acting on a body.

5 0
3 years ago
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 c
valina [46]

Answer:

q_1 =7.08*10^{-9}C.

Explanation:

Gauss's Law says that the electric flux \Phi_E through a closed surface is directly proportional to the charge Q_{enc} inside it. More precisely,

$\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}. $

This means what is outside this closed surface S does not contribute to the flux through it because field lines that go in must come out, <em>resulting a zero flux from an external charge. </em>

In our context, this means the charge q_2 which is outside the sphere will have zero flux through the surface; therefore, Gauss's law will only be concerned with charge q_1 which is inside the sphere; Hence,

$\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C. $

Solving for q_1 gives

$ q_1= (800 N\cdot m^2/C)\epsilon_0, $

$ q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2) $

\boxed{q_1 =7.08*10^{-9}C. }

which is the charge inside the sphere.

5 0
3 years ago
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