Particles in a ______<br> have a small amount of energy, vibrate, and are stuck in place?

Identify the activities that help reduce the carbon footprint of a person or an organization

fiasKO [112]

All pf them except for the train.

3 0

3 years ago

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

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Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy

Anestetic [448]

Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

$K.E = \frac{1}{2}mv^{2}$

where,

K.E = Kinetic Energy

m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

$K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}$

<u>Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.</u>

4 0

3 years ago

If u accomplished 10,000 newton meters of work how much work did you do in units of joules

Amiraneli [1.4K]

1.0 joule= 1.0 newtons × 1.0 meter = 1.0 newton × meter

Work = 10 newtons × 5 meters = 50 newton × meter

3 0

4 years ago

What challenges are faced by scientist investigating effects of radiation in Chernobyl and Eniwetok Atoll in the Chernobyl accid

Andrew [12]

Answer:

The health hazard of directly going to these places to take direct readings and observations.

Explanation:

The Chernobyl accident was a catastrophic radioactive accidents, with an immediate fatal effect on some victims (mostly firemen and law enforcement officers). Scientists investigating the effects of the radiation left on the venues of the accident face a health risk if they intend to carry out a full scale on-field testing and experiment. Some of the relocated occupants of these places, that have long been relocated, have shown some health effect of the radiation. And scientist studying and working on these places for too long stand a risk of developing health complications if they are exposed to the radiations for too long.

8 0

4 years ago

Particles in a ______ have a small amount of energy, vibrate, and are stuck in place?