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Vlada [557]
2 years ago
14

the car turned a corner going 10m/s if it took two seconds to turn what is its acceleration is this acceleration​

Physics
1 answer:
exis [7]2 years ago
6 0

The acceleration of this car is equal to 5 m/s^2.

<u>Given the following data:</u>

  • Initial velocity = 0 m/s (assuming it's starting from rest).
  • Final velocity = 10 m/s.
  • Time = 2 seconds.

To determine the acceleration of this car:

<h3>How to calculate acceleration.</h3>

In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.

Mathematically,  acceleration is given by this formula:

a = \frac{V\;-\;U}{t}

<u>Where:</u>  

  • V is the final velocity.
  • U is the initial velocity.
  • is the time measured in seconds.

Substituting the given parameters into the formula, we have;

a = \frac{10\;-\;0}{2}

Acceleration, a = 5 m/s^2

Read more on acceleration here: brainly.com/question/24728358

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The relationship between body mass body weight and body density
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The formula for calculating Density is:

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From this formula, we can say that the relationship between Mass and Density is a direct one. In other words, if mass is increasing - all else being equal - then density will increase as well.

If mass however was decreasing, density would have to decrease as well.

For example, assume 3 bricks have masses of 5kg, 10kg and 15kg. Also assume that the bricks all have the same volume of 5 m³.

Density of 5kg brick = 5 / 5 = 1 kg/m³

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Un vagón de 250 kg situado en la cima de una montaña rusa inicia su descenso por una rampa inclinado 60º sobre la horizontal. Si
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3 years ago
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at
VARVARA [1.3K]

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

6 0
4 years ago
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