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2 years ago

A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?

Physics

1 answer:

balandron [24]2 years ago

7 0

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s Result

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1 year ago

A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15

cupoosta [38]

Answer:

$\boxed {\boxed {\sf 4.5 \ m/s \ in \ the \ positive \ direction}}$

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

$v_f= v_i + at$

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

$v_i$ = 2.7 m/s
a= 0.15 m/s²
t= 12 s

Substitute the values into the formula.

$v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)$

Multiply the numbers in parentheses.

$v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)$

$v_f = 2.7 \ m/s + (0.15 \ m/s *12)$

$\v_f=2.7 \ m/s + (1.8 \ m/s)$ $v_f=2.7 \ m/s + (1.8 \ m/s)$

Add.

$v_f=4.5 \ m/s$

The final velocity of the boat is 4.5 meters per second in the positive direction.

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2 years ago