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4 years ago

The stored energy in a piece of coal can be _________ into heat energy.

Physics

2 answers:

Leviafan [203]4 years ago

7 0

The answer is transformed
because you changed the kind of energy

gavmur [86]4 years ago

4 0

The Law of Conservation of Energy states that energy can either be transferred from one object to another, or transformed from one type to another, but never destroyed. The stored energy in coal is chemical energy because it is ready to react, and when the coal is burned, the chemical energy is transformed by the reaction into heat energy.

Answer: a. transformed

Hope this helps!

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Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0

3 years ago

During energy transformation, all energy systems: are

Tom [10]

B: Energy lose

i say this because in order to change they lose energy.

3 0

3 years ago

Read 2 more answers

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the

Margaret [11]

Answer:

The change in temperature, $\Delta T=9.45^{\circ} C$

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from $-4^{\circ} F\ to\ 45^{\circ} F$ in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,