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Katen [24]
3 years ago
10

You must exert a force of 4.5 N on a book to slide it across a table. If you do 2.7J

Physics
1 answer:
hoa [83]3 years ago
3 0

Answer: 0.6m

Explanation:

Given that:

force = 4.5 N

Work done = 2.7J

Distance moved by the book = ?

Since work is done when force is applied on an object over a distance, apply the formula:

work = force x distance

2.7J = 4.5N x distance

Distance = (2.7J / 4.5N)

Distance = 0.6 m

Thus, the book was moved 0.6 metres far

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A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample
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2 years ago
Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

  • V = Linear speed
  • W = Angular speed
  • r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2\pi/T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x 10^{-4} Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x  10^{-4} x 149.6 x 10^{6}

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4\pi ^{2}r^{3}) / GT^{2}

M = (4 x 9.8696 x 3.35 x 10^{24}) / (6.67 x 10^{-11} x 7.68 x 10^{11}<em>)</em>

<em>M = 1.32 x </em>10^{26} / 51.226

M = 2.58 x 10^{24} Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

#SPJ1

3 0
1 year ago
Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner
stepan [7]

Answer:

   P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          A₁ v₁ = A₂ v₂

Let's look for the areas

          r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

          r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

          A₁ = π r₁²

          A₁ = π 1.125²

          A₁ = 3,976 cm²

          A₂ = π r₂²

          A₂ = π 0.1²

          A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

           v₁ = v₂ A₂ / A₁

           v₁ = 11.2 0.0452 / 3.976

           v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

           P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

          P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

           P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

           P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

           P₁ = 2.3506 10⁵ Pa

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3 years ago
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