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Sauron [17]
3 years ago
9

Advantages of cutting crops with sickle ? please friends give this answer in easy way....​

Physics
1 answer:
Eva8 [605]3 years ago
6 0

Answer: The only advantage cutting crops with a sickle is that its very effective for cutting lodged crops.

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What do electrons move from
Alona [7]

Answer:

Negatively charged, to positively charged parts

Explanation:

Electrons are negative, negative is attracted to positive.

7 0
3 years ago
Which of the following is NOT common of elite Shang burials?
Zigmanuir [339]

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

6 0
2 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 diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignit
Klio2033 [76]

Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1}  = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given:  T₁= 18°C we convert to  Kelvin(K) by adding 273.

∴      T₁= 18°C + 273 = 291K

        T₂ = 733° C  also,we convert to  Kelvin(K) by adding 273

 ∴      T₂ = 733° C +273 =1046K

          γ = 7/5

 ∴ Rearranging equation(1), we have

    (T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

   (V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

    V₁/V₂ = 24.42.

∴ Compression ratio = 24.42

7 0
3 years ago
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh
ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

  • Assuming no external forces acting during the collision, total momentum must be conserved.
  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       p_{o1x} = 2.00 kg*m/s (1)

       p_{o1y} = 0 (2)

  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)

  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

  • We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)

  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

  • Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

6 0
2 years ago
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