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Elanso [62]
3 years ago
5

Find the weight of an object of mass 5 kg

Physics
2 answers:
san4es73 [151]3 years ago
4 0

Answer:

1

For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is 1.67 m/s2, the object weighs 8.4 N.

2

The weight of the object on the surface of earth would be 5kg. The weight of the object on the surface of moon would be 5/6= 0.83 as the weight of any object on earth is six times than that on moon.

Elis [28]3 years ago
3 0

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

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The element is iridium and it has 77 electrons

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A student places two books on a table. One book weighs less than the other book
Gnoma [55]
The book that weighs less
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A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

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4 0
1 year ago
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse
777dan777 [17]

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

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      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

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Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

8 0
3 years ago
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anastassius [24]

Answer:

plastic isn't conductive

Explanation:

Correct me if I'm wrong but plastic isn't conductive copper is so if it's not conductive it will probably mess up the circuit or somthing

4 0
2 years ago
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