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3 years ago

Is the box moving at a constant speed? Explain how you know. What does this tell you about the kinetic energy of the system

Physics

1 answer:

mars1129 [50]3 years ago

8 0

Answer:

The box is moved at constant speed, then the change in kinetic energy is zero.

Explanation:

Let suppose that box moves at constant speed, meaning that its kinetic energy ( $K$ ), measured in joules, is expressed by the following equation:

$K = \frac{1}{2}\cdot m\cdot v^{2}$ (1)

Where:

$m$ - Mass, measured in kilograms.

$v$ - Speed, measured in meters per second.

If the box moves at constant speed, then we notice that $v_{f}^{2}-v_{o}^{2} = 0$ , therefore, $\Delta K = 0$ . In a nutshell, if the box is moved at constant speed, then the change in kinetic energy is zero.

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A 140 g baseball is moving horizontally to the right at 35 mis when it is hit by the bat. the ball fl ies off to the le ft at 55

Anit [1.1K]

Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

When it hits the bat, it flies left with a velocity of 55 m/s at an angle of 25°. To reiterate:

mass = 0.140kg, Initial: 35m/s, Final: 55m/s at an angle of 25°

For this problem, we have to use the impulse equation, but before that lets break the velocity into components (It will be apparent towards the end):

The initial velocity is moving only in the horizontal direction, so:

$v_{0x} = 35 m/s$

The final velocity has an x and a y component:

$v_{fx} = 55cos(25) = 49.84692829m/s\\v_{fy} = 55sin(25) = 23.2440044m/s$

Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

$J = dp= m*dv$

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

$dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s$

Finally, substitute into the equation

$J = m*dv\\J = 0.140kg * 87.97320604m/s\\J = 12.31624885kg*m/s\\J = 12.32 kg*m/s$

J = 12.32kg*m/s

8 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it

White raven [17]

Answer:

The positions are 0.0194 m and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

$v = \omega \sqrt{A^2-x^2}$

the maximum speed of the simple harmonic motion is given as;

$v_{max} = \omega A$

when the speed equal one fourth of its maximum speed

$v =\frac{v_{max}}{4}$

$\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m$