Physics, 06.01.2021 20:20 arisvianeygd3

How fast is a 30 kg child runnning if her kinetic energy is 500 J?

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Answer from: sophiqdionacles1968

$\displaystyle v = 5.7735 \ m/s$

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

Physics

SI Units

Speed is m/s (meters per second)Energy is J (Joules), equal to 1 N/m (newton per meter)Mass is kg (kilograms)

Energy

Kinetic Energy Formula: $\displaystyle KE = \frac{1}{2} mv^2$

KE is Kinetic Energy (in J)m is mass (in kg)v is velocity (in m/s)Explanation:

Step 1: Define

Mass m = 30 kg

Kinetic Energy KE = 500 J

Step 2: Solve for v

Substitute [KEF]: $\displaystyle 500 = \frac{1}{2}(30)v^2$ Multiply: $\displaystyle 500 = 15v^2$ Isolate v term: $\displaystyle \frac{100}{3} = v^2$ Isolate v: $\displaystyle \frac{10\sqrt{3} }{3} = v$ Rewrite: $\displaystyle v = \frac{10\sqrt{3} }{3} \ m/s$ Evaluate: $\displaystyle v = 5.7735 \ m/s$

Answer from: Quest

answer; ///i believe that the correct answer is (45*) at a 45 degree check first good

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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