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Basile [38]
3 years ago
13

Based on the free-body diagram, the net force acting on this firework is

Physics
2 answers:
Strike441 [17]3 years ago
4 0

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

Gnom [1K]3 years ago
4 0

Answer:

0 on edge !!!

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What are the conditions of equilibrum
natta225 [31]

Answer:

they are given below

The sum or resultant of all external forces acting on the body must be equal to zero.

The sum or resultant of all external torques from external forces acting on the object must be zero.

3 0
3 years ago
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"On a good day, it takes Mr. Hess 23 minutes (1380 seconds) to drive the 12 miles (19,300 meters). What is his average velocity
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Answer:

his average velocity is

v =  \frac{19300}{1380}  = 13.9855(m \div s)

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2 years ago
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

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A load of bricks is lifted to the second floor of a building. How do work and power relate to this job?
Nady [450]
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Myopia is a condition of the eye where the crystalline lens focuses the light rays to a position between the lens and the retina
aliya0001 [1]

Answer: concave lens

Explanation:

Myopia is a condition of the eye where someone can only see near distant object clearly but not far distant object.

Myopia is corrected using concave lens (diverging) in order to diverge the rays entering the eye thereby allowing the rays to be focused properly on the retina.

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