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rewona [7]
3 years ago
8

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

answer to two significant figures and include the appropriate units?
Physics
1 answer:
torisob [31]3 years ago
6 0
<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R 
I = MR² = (0.200)(5) = 1.00 kg-m² 
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal 
V = 3.0 m/s 
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s 
w = [V component / R] = 1.3433/√5 = 0.601 rad/s 
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>


i hope I'm right</span>
You might be interested in
How many coulombs of positive charge are there in 47.0 gm of plutonium, given its atomic mass is 244 and that each plutonium ato
enot [183]

Answer:

  • 1.78×10⁶ C

Explanation:

Using the atomic mass of pluonium atoms (244 g/mol), you can calculate the number of atoms in 47.0 g. Then, knowing that each plutonium atom has 96 protons, you calculate the number of protons in the 47.0 g sample. Finally, using the positive charge of one proton, you calculate the total positive charge in the 47.0 g of plutonium.

<u>1. Number of atoms of plutonium in 47.0 g</u>

  • Number of moles = mass / atomic mass = 47.0 g / 244 = 0.1926 moles

  • Number of atoms = number of moles × 6.022 × 10²³ atoms/mol

  • Number of atoms = 0.1926 mol × 6.022 × 10²³ atoms/mol = 1.15998×10²³ atoms

<u>2. Number of protons</u>

  • Number of protons = 1.15998×10²³ atoms × 96 protons/atom = 1.11385×10²⁵ protons

<u>3. Charge</u>

<u />

  • Charge = charge of one proton × number of protons

  • Charge = 1.602×10⁻¹⁹ C/proton × 1.11385×10²⁵ protons = 1.78×10⁶C
7 0
3 years ago
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
BartSMP [9]

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
Which statement is correct?
slega [8]

Answer:B

Explanation:

Galilean transformation are only approximately correct,while Lorentz transformation are more exact

6 0
3 years ago
Read 2 more answers
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
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