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SSSSS [86.1K]
2 years ago
8

A 1.72-kg block of soft wood is suspended by two strings from the ceiling. The wood is free to rotate in pendulum-like fashion w

hen a force is exerted upon it. A 8.50-g bullet is fired into the wood. The bullet enters the wood at 431 m/s and exits the opposite side shortly thereafter. If the wood rises to a height of 13.8 cm, then what is the exit speed of the bullet?
Physics
1 answer:
romanna [79]2 years ago
4 0

Answer:

Vf_{2} = 98.21 m/s                    

Explanation:

The block has a final potential energy

U = mgh

U = (1.72)(9.8)(0.138)

    = 2.3261 J

How after of the collision the block's mechanical energy is conserved, then we can calculate the block's initial velocity

     2.3261 = \frac{mv^{2} }{2}

     2.3261 = \frac{1.72v^{2} }{2}

            v = 1.6446 m/s

In the collision is conserved the lineal momentum of the system then:

m_{1}Vo_{1} + m_{2}Vo_{2} = m_{1}Vf_{1} + m_{2}Vf_{2}

(1.72)(0) + (0.0085)(431) = (1.72)(1.6446) + (0.085)Vf_{2}

3.6635 = 2.8287 + 0.0085Vf_{2}

Vf_{2} = 98.21 m/s                    

 

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It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0
VikaD [51]
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have 

rate = \frac{-(C_{new} - C_{old})}{t}

where C are the concentration values and t is the time taken for it to decompose.

14 = \frac{-(0.02 - 0.06)}{t}
t = 0.003 s

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0
3 years ago
Nepal has high potential for producing hydroelectricity however it is difficult too.
Stels [109]

Answer:

I'm not a scholar of hydroelectric power in Nepal, so consider my answers carefully, below.

Explanation:

High Potential:  Hydroelectric power comes from the potential energy stored in a mass that is above Earth's surface.  As the word "hydro" implies, the mass in this case is water.  Water from snow and glacier melt, and from normal precipitation (rain) in mountainous regions eventually cascades down the mountains in fast-flowing rivers or waterfalls.  Often, there are lakes or man-made reservoirs to collect and store the water before it flows down.  Mt. Everest is 8848 meters tall (about 29,000 feet).    If a lake forms at just 2,000 meters, one can calculate the amount of energy in each kilogram of water stored in the lake that represents the potential energy available at that altitude.  1 kg of water at 2,000 meters has potential energy, PE, according to the equation:  PE = mgh, where m is the mass in kg, g is Earth's acceleration due to gravity (9.8 m/sec^2), and h is height, in meters.  

PE = mGH

PE = (1 kg)*(9.8 m/sec^2)*(2,000 meters) = 19,000 kg*m/sec^2

1 kg*m/sec^2 is the SI unit for 1 Joule, a measure of energy.

This potential energy can be converted into electrical energy by releasing the water so that it can flow down to a water-powered turbine that spins magnets and coils of wire that produce electricity.  The 19,000 Joules of water potential energy can be converted to electrical power, less any inefficiency in the system, such as friction.

Nepal has the natural advantage in that it has many high mountain ranges with water flows that can be used for generating electrical power.  The result is low operating costs (the fuel is the flowing water) and no greenhouse gas emissions

The difficulty in developing hydroelectric power in Nepal is due to the same factor that gives it an advantage:  it is difficult constructing large hydroelectric plants in such rough terrain, and the power lines that are needed to transport the power to its destination are expensive and difficult to maintain and repair.

8 0
2 years ago
For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or
trapecia [35]

Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.

7 0
3 years ago
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
dedylja [7]
<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.

\theta=30\° the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at 30\°:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))   (2)

\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m   (3)

Now, the initial energy E_{o}=400keV=400(10)^{3}eV of the photon is given by:

 E_{o}=\frac{h.c}{\lambda_{o}}    (4)

From this equation (4) we can find the value of \lambda_{o}:

\lambda_{o}=\frac{h.c}{E_{o}}    (5)

\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}    

\lambda_{o}=3.102(10)^{-12}m    (6)

Knowing the value of \Delta \lambda and \lambda_{o}, let's find \lambda':

\Delta \lambda=\lambda' - \lambda_{o}

Then:

\lambda'=\Delta \lambda+\lambda_{o}  (7)

\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m  

\lambda'=3.427(10)^{-12}m  (8)

Knowing the wavelength of the scattered photon \lambda'  , we can find its energy E' :

E'=\frac{h.c}{\lambda'}    (9)

E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}    

E'=362.063keV    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron E_{e}, we have to calculate all the energy lost by the photon, which is:

E_{e}=E_{o}-E'  (11)

E_{e}=400keV-362.063keV  

Finally we obtain the energy of the recoiling electron:

E_{e}=37.937keV  

5 0
3 years ago
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krok68 [10]

Answer:

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<h2>=> </h2>

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