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joja [24]
2 years ago
13

What kind of energy transformation occurs when you lite a firecracker

Physics
2 answers:
Cerrena [4.2K]2 years ago
7 0
The answer is chemical energy.
d1i1m1o1n [39]2 years ago
5 0
Friction=Mechanical turn into Thermal
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Describe a situation when you might travel at a high velocity bit with low acceleration
iren [92.7K]

I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy.  They start to close.
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and I have just finished my dinner.
Finally I can't help it.  Resistance is futile.
I give up, and fall deep asleep.
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My seat is in row 26 on the airplane I'm flying in
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5 0
3 years ago
The cycle of the moon through its phases or the synodic month is___.
ikadub [295]
It is 29 and a half days long
3 0
3 years ago
Read 2 more answers
The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid
timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0
3 years ago
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
siniylev [52]

Answer:

The height of the  object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of Gmm_{e}

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

200=\dfrac{Gmm_{e}}{(3958.756)^2}

200\times(3958.756)^2=Gmm_{e}

Gmm_{e}=3.134\times10^{9}

We need to calculate the height of the  object

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

125=\dfrac{200\times(3958.756)^2}{r^2}

r^2=\dfrac{200\times(3958.756)^2}{125}

r^2=25074798.5

r=\sqrt{25074798.5}

r=5007.4\ miles

Hence. The height of the  object is 5007.4 miles.

7 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
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