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vesna_86 [32]
2 years ago
10

What is the molecular formula for the hydrocarbon in the mass spectrum above?

Physics
1 answer:
PtichkaEL [24]2 years ago
6 0
What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following $1 \mathrm{H}$ NMR spectrum? 7.25 $\delta(5 \mathrm{H}, \text { broad singlet); } 2.90 \delta(1 \mathrm{H}, \text { septet, } J=7 \mathrm{Hz}) ; 1.22 \delta(6 \mathrm{H},\text { doublet, }$ $J=7 \mathrm{Hz})$
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Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are
Mars2501 [29]

Answer:

 θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

F = \sqrt{F_1^2 + F_2^2}

F = \sqrt{22.8^2 + 16.6^2}

F = \sqrt{795.4}

      F = 28.20 N

direction

\theta = tan^{-1}(\dfrac{F_2}{F_1})

\theta = tan^{-1}(\dfrac{16.6}{22.8})

\theta = tan^{-1}(0.728)

       θ = 36°

5 0
3 years ago
How is energy transferred throughout the roller coaster
solmaris [256]

Answer:

On a roller coaster, energy changes from potential to kinetic and back again many times over and over the course of the ride. Kinetic energy is energy that an object has because of its motion. All moving objects possess kinetic energy, which is determined by the mass and speed of the object.

Explanation:

3 0
3 years ago
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Does mercury have more protons and electrons than tin
Klio2033 [76]
Yes it does.
On the periodic table, tin is #50 and Mercury is # 80. 
3 0
3 years ago
What may result when the energy that builds up at plate boundaries is released because the plates suddenly overcome the force of
Arlecino [84]
It has to be Earthquake
4 0
2 years ago
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The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
2 years ago
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