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Mars2501 [29]
3 years ago
10

A certain metal wire has a cross sectional area of 1 mm2 and is 1 m long. when it is hung from the ceiling and a 10 kg mass is h

ung from the bottom of it, the wire stretches 1 mm. treat the wire as a single macroscopic spring. what is its effective spring constant?
Physics
1 answer:
kvasek [131]3 years ago
5 0
From the Hooke's law , the extension force of an elastic material is directly proportional to the extension. 
That is, F = k e, where F is the force , k is the constant and e is the extension
 F = 10 × 10 = 100 N
e = 1mm or 0.001 m
Hence, k = F/e
                = 100 N/ 0.001
                = 100000 N/m or 100 N/mm
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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in
Lana71 [14]

Answer:

The average power is calculated as 735.0 W

Solution:

As per the question:

Total mass, M = 1200 kg

Counter mass of the elevator, m = 950

Distance traveled by the elevator, d = 54 m

Time taken, t = 3 min = 180 s

Now,

To calculate the average power:

First, we find the force needed for lifting the weight:

Force, F = (M - m)g = (1200 - 950)\times 9.8 = 2450 N

Now, the work done by this force:

W = Fd = 2450\times 54 = 132300\ J = 132.3\ kJ

Average power is given as:

P_{avg} = \frac{W}{t} = \frac{132300}{180} = 735.0\ W

4 0
3 years ago
Which type of cloud would you expect to be involved in some form of precipitation?
fomenos
Stratus clouds maybe
6 0
3 years ago
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.
Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] =  11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

7 0
3 years ago
Which telescope would be better viewing a faint, distant star? Why?
grin007 [14]
Reflecting telescope. Reflecting telescopes tend to have larger objective (due to the use of mirrors, mirrors are a lot cheaper than lenses) and have the ability to collect more light, while refracting telescopes are limited to objective lenses with smaller diameters due to their structural limitations (chromatic abbreviation, for example). Therefore, reflecting telescopes should be better at viewing faint distant stars
4 0
3 years ago
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