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3 years ago

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How would you calculate an object’s mechanical energy? a. Add its kinetic and potential energies. b. Multiply its kinetic and po

tential energies. c. Subtract its kinetic energy from its potential energy. d. Subtract its potential energy from its kinetic energy.

2 answers:

dalvyx [7]3 years ago

6 0

A. Add it's Kinetic and Potential energies

Ahat [919]3 years ago

6 0

We know that:
EM=Ek+Epg
So basically you just sum the kinetic plus the potential energy and that´s it. (a)

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lawyer [7]

The bigger the object the greater the gravitational pull, so the farther away the big object is its gravitational force begins to decrease. Refer to the picture for more explanation.

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2 years ago

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

vovangra [49]

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

Momentum = mass x velocity

Since the speed of the ball is unknown:

Velocity = $\frac{momentum }{mass}$

= $\frac{0.78}{0.15}$

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2 years ago

A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum

Kryger [21]

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

$v_{\max }=\sqrt[]{2gh}$

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

$\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}$

Hence, the maximum speed of the pendulum is 2.42 m/s

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7 months ago

What atmospheric gas is common in the outer planets but rare in the inner planets

Degger [83]

Answer:

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Explanation:

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3 years ago

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro

LiRa [457]

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$ , $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$ )

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$

3 0

2 years ago