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3 years ago

Which objects have both potential and kinetic energy? Check all that apply.

Physics

2 answers:

wariber [46]3 years ago

3 0

A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.

denis23 [38]3 years ago

3 0

Answer:

a sky diver who is midway through his descent to the ground

a roller coaster car halfway down the second hill

a falling raindrop

Explanation:

From the law of conservation of energy, it can neither be created nor destroyed but can be converted from one form to another.

The sum of kinetic energy and potential energy is mechanical energy. A body has potential energy due to its position or configuration and kinetic energy due to its speed.

A body can have both potential energy and kinetic energy.

The correct options are:

a sky diver who is midway through his descent to the ground
a roller coaster car halfway down the second hill
a falling raindrop

A boulder resting on top of a mountain has just potential energy.

A parked car also has just potential energy.

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For a mass oscillating on a spring at what positions are (a) velocity and (b) acceleration of the mass have maximum valeus?

EleoNora [17]

Answer:

a)At the mean position

b)At the extremes positions

Explanation:

Given that mass is having oscillation motion.

We know that

1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.

2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.

Therefore

a)At the mean position

b)At the extremes positions

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3 years ago

A moving object is in equilibrium. Which best describes the motion of the object if no forces change?

Alchen [17]

If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.

8 0

3 years ago

Read 2 more answers

Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t

andre [41]

Answer:

2.5 s, 5 m

Explanation:

The equations for the horizontal and vertical position of Lukalu are:

$x(t) = 8t\\y(t) = -16t^2 + 100$

we can find the time it takes her to reach the ground by requiring that the vertical position becomes zero:

y(t) = 0

So we find:

$0=-16t^2 +100\\16t^2 = 100\\t=\sqrt{\frac{100}{16}}=2.5 s$

The horizontal distance of Lukalu instead will be given by the equation for the horizontal position, substituting t = 2.5 s:

$x=8t = 8 \cdot 2.5 s =5 m$

4 0

3 years ago

Determine the capacitive reactance for a 20 uF capacitor that is across a 20 volt, 60 Hz source

aleksandr82 [10.1K]

Answer:

Capacitive reactance is 132.6 Ω.

Explanation:

It is given that,

Capacitance, $C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F$

Voltage source, V = 20 volt

Frequency of source, f = 60 Hz

We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :

$X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}$

$X_C=132.6\ \Omega$

So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.

4 0

3 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

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3 years ago