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kumpel [21]
3 years ago
6

a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?

Physics
1 answer:
BartSMP [9]3 years ago
5 0

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

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Nuclear power plants, wind farms, water farms, and geothermal heating
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A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
Virty [35]

Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
  • g= 9.8 m/s²
  • h= 20 m

Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

7 0
3 years ago
Calculate the mass of a liquid with a density of 2.5 g/ml and a volume of 15ml
Nana76 [90]
Using the density equation and clearing mass:

\rho = \frac{m}{V}\ \to\ m = \rho\cdot V = 2.5\frac{g}{mL}\cdot 15\ mL = \bf 37.5\ g
3 0
3 years ago
Why are weight and mass used synonymously on earth?
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Weight and mass are used synonymously on earth because the value of g is constant on the earth because the weight of a body is the amount of force exerted by the Earths gravity on an object of finite mass. And, the mass of an object gives the amount of matter in the body and is measured in kilograms. 
6 0
3 years 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

7 0
3 years ago
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