Explanation:
Efficiency = work out / work in
e = 571 J / 1694 J
e = 0.337
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
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<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.
<u>Explanation:
</u>
Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.
As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.
Answer:Speed can be defined as
Explanation: B because if you think about it, 50 milers per hour means they are dividing the time it takes to drive a certain distance going that speed. So id has to be answer B! your welcome. Please upvote only if its correct!
Answer:
Lauch velocity (u) = 26.15 m/s
Lauch Angle (θ) = 35°
Explanation:
From the question given above, the following data were obtained:
Range (R) = 65 m
Time of flight (T) = 3 s
Acceleration due to gravity (g) = 10 m/s²
Lauch velocity (u) =?
Lauch Angle (θ) =?
R = u²Sin2θ /g
65 = u² × Sin2θ /10
Recall:
Sin2θ = 2SinθCosθ
65 = u² × 2SinθCosθ / 10
65 = u² × SinθCosθ / 5
Cross multiply
65 × 5 = u² × SinθCosθ
325 = u² × SinθCosθ .....(1)
T = 2uSinθ / g
3 = 2uSinθ / 10
3 = uSinθ / 5
Cross multiply
3 × 5 = uSinθ
15 = u × Sinθ
Divide both side by Sinθ
u = 15 / Sinθ....... (2)
Substitute the value of u in equation (2) into equation (1)
325 = u² × SinθCosθ
u = 15 / Sinθ
325 = (15 / Sinθ)² × SinθCosθ
325 = 225 / Sin²θ × SinθCosθ
325 = 225 × SinθCosθ / Sin²θ
325 = 225 × Cosθ / Sinθ
Cross multiply
325 × Sineθ = 225 × Cosθ
Divide both side by Cosθ
325 × Sineθ / Cosθ = 225
Divide both side by 325
Sineθ / Cosθ = 225 / 325
Sineθ / Cosθ = 0.6923
Recall:
Sineθ / Cosθ = Tanθ
Tanθ = 0.6923
Take the inverse of Tan
θ = Tan¯¹ 0.6923
θ = 35°
Substitute the value of θ into equation (2) to obtain the value of u.
u = 15 / Sinθ
θ = 35°
u = 15 / Sin 35
u = 15 / 0.5736
u = 26.15 m/s
Summary:
Lauch velocity (u) = 26.15 m/s
Lauch Angle (θ) = 35°
