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Vlad [161]
3 years ago
7

The way to earn a college diploma is by?

Physics
2 answers:
Dima020 [189]3 years ago
8 0
Get good marks in college
Morgarella [4.7K]3 years ago
4 0
Going to college and passing all your classes
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In which graph is acceleration the slope?
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Answer:

<em>Velocity</em><em> </em><em>-</em><em>time</em><em> </em><em>graph</em><em> </em>

Explanation:

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4 0
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The largest stars are _____ times the mass of the Sun. 10 100 1,000 10,000
Arte-miy333 [17]
<h3><u>Answer;</u></h3>

100 times

<h3><u>Explanation;</u></h3>
  • The largest stars are 100 times the mass of the Sun.
  • <u>The giant stars are about 10 to 100 times the radius of the sun</u>, which means they are 100 times brighter than the sun.
  • <em><u>The largest known star in terms of mass and brightness is known as the Pistol Star. It is believed to be 100 times as massive as our Sun, and 10,000,000 times as bright.</u></em>
5 0
3 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

4 0
3 years ago
An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti
Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

  • There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
  • So, we can write the following expression for Fg:

       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

  • The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

  • According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

       a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2  (3)

3 0
3 years ago
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