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Alla [95]
2 years ago
10

A projectile is launched with an initial speed of 40 mys from the foor of a tunnel whose height is 30 m. What angle of elevation

should be used to achieve the maximum possible horizontal range of the projectile?
Physics
1 answer:
KiRa [710]2 years ago
7 0

Answer:

We are given the trajectory of a projectile:

y=H+xtan(θ)−g2u2x2(1+tan2(θ)),

where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then

H+xtan(θ)−12Lx2(1+tan2(θ))=0

Differentiate both sides with respect to θ.

dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0

Solving for dxdθ yields

dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))

This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.

H+L−12Lx2−L2=0.

This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.

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A 0.614mole sample of ideal gas at 12degree occupies a volume of 4.32.what is the pressure of the gas
Kaylis [27]

Answer:

336.9520 atm

Explanation:

The Gas Equation is as follows;-

Pressure×Volume=Number of Moles × Universal Gas Constant ×Temperature(in Kelvin)

Given Parameters

Number of moles-0.614 mol

Temperature 12°C or 12+273.15 ie 285.15°F

Volume-4.32 L

Universal Gas Constant-8.314 J/mol·K

Pressure -?(in atm)

Plugging in all the values in the Gas Equation:-

Pressure=\frac{0.614 × 8.314× 285.15}{4.32}   atm

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3 0
3 years ago
A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
goldfiish [28.3K]

Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

Now, let us verify the above result without any conversion.

Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

8 0
3 years ago
Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, t
user100 [1]

Answer:

The width of the strand of hair is   1.96 10⁻⁵ m

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For this diffraction problem they tell us that it is equivalent to the diffraction of a single slit, which is explained by the equation

<h3>       a sin θ =±  m λ </h3><h3 />

Where the different temrs are: “a” the width of the hair, λ the wavelength, θ the angle from the center, m the order of diffraction, which is the number of bright rings (constructive diffraction)  

We can see that the diffraction angle is missing, but we can find it by trigonometry, where L is the distance of the strand of hair to the observation screen and "y" is the perpendicular distance to the first minimum of intensity

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      Tan θ = y/L

      Tan θ = 5.06/125

      θ = tan⁻¹ ( 0.0405)

      θ =  2.32º

With this data we can continue analyzing the problem, they indicate that they measure the distance to the first dark strip, thus m = 1

     a = m λ / sin θ

     a = 1 633 10⁻⁹ 1.25/sin 2.3

     a = 1.96 10⁻⁵ m  

     a = 0.0196 mm

The width of the strand of hair is   1.96 10⁻⁵ m

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Answer:

C

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