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bija089 [108]
3 years ago
15

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1. 2 m high. Student 2 pushes an identical box up a 5 m ra

mp from the floor to the top of the same table. Which student did the MOST work?.
Physics
1 answer:
valina [46]3 years ago
5 0

The student 2 did the MOST work because he pushed the box to a greater height.

The given parameters:

  • <em>Force applied by Student 1 = 500 N</em>
  • <em>Height of table for student 1 = 1.2 m</em>
  • <em>Height of ramp for student 2 = 5 m</em>

The work done by student 1 is calculated as follows;

Work = P.E = mgh

P.E = 1.2mg

The work done by the student 2 is calculated as;

P.E = 5mg

where;

  • <em>m </em><em>is the </em><em>mass </em><em>of the box</em>

Thus, we can conclude that the student 2 did the MOST work.

Learn more about work done here: brainly.com/question/8119756

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Are small molecular units joined together in large molecules
inysia [295]
<span>Polymers are small molecular units joined together in large molecules.

hope this helps!</span>
7 0
3 years ago
Plz help This is for Rock formations...only 5 sentences if not more
dsp73

Answer:

The coastal zone is not a stable and constant environment, but a dynamic place that can change rapidly in response to natural processes such as seasonal weather patterns. Waves, winds, currents, tides and storms are the major forces on the coast.

Explanation:

8 0
3 years ago
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
Oxana [17]

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0
3 years ago
An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

                                                                     = 780/130

                                                                     = 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

                                                                                          = 910 J

Learn more about  coefficient here:

brainly.com/question/18915846

#SPJ4

5 0
2 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
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