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3 years ago

Does the KE of a car change more when it goes from 10 to 20 km/h or when it goes from 20 to 30 km/h?

Mathematics

1 answer:

Ludmilka [50]3 years ago

5 0

<h2>The change in KE is more for speed change from 20 to 30 km/h</h2>

Step-by-step explanation:

Let the mass of car be m.

Speed change from 10 to 20 km/h.

Initial speed = 10 km/hr

Initial KE = 0.5 x m x 10² = 50 m

Final speed = 20 km/hr

Final KE = 0.5 x m x 20² = 200 m

Change in KE = 200m - 50m = 150m

Speed change from 20 to 30 km/h.

Initial speed = 20 km/hr

Initial KE = 0.5 x m x 20² = 150 m

Final speed = 30 km/hr

Final KE = 0.5 x m x 30² = 450 m

Change in KE = 450m - 200m = 250m

The change in KE is more for speed change from 20 to 30 km/h

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A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long

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Answer: 1.4 seconds

Step-by-step explanation:

The equation is: h(t) = at² + v₀t + h₀ where

a is the acceleration (in this case it is gravity)
v₀ is the initial velocity
h₀ is the initial height

Given:

a = -9.81 (if it wasn't given in your textbook, you can look it up)
v₀ = 12
h₀ = 3

Since we are trying to find out when it lands on the ground, h(t) = 0

EQUATION: 0 = 9.81t² + 12t + 3

Use the quadratic equation to find the x-intercepts

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$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\$

Note: Negative time (-0.2) is not valid

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3 years ago

Nico Joven presently earns $348.40 per week. He claims

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Answer:

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Answer:

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Step-by-step explanation:

$42\div60$

Calculate

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Cross out the common factor

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Multiply a number to both the numerator and the denominator

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Write as a single fraction

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I hope this helps you

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Answer:

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Step-by-step explanation:

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