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kap26 [50]
3 years ago
12

A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt

+βt3θ(t)=γt+βt3, where γ=γ= 0.450 rad/srad/s and β=β= 1.10×10−2 rad/s3rad/s3. Part A Calculate the angular velocity of the merry-go-round as a function of time. Express your answer in terms of the variables ββ, γγ, and tt. ωz(t)ωz(t) = nothing rad/srad/s SubmitRequest Answer Part B What is the initial value of the angular velocity? ωzωz = nothing rad/srad/s SubmitRequest Answer Part C Calculate the instantaneous value of the angular velocity ωzωz at t=t= 5.25 ss . ωzωz = nothing rad/srad/s SubmitRequest Answer Part D Calculate the average angular velocity ωav−zωav−z for the time interval t=0t=0 to t=t= 5.25 ss .
Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

hufiui

fihgpfghlfikgergkfkjhfkhjgkffhhh

Explanation:

jjgzgcjxhgygueyuufhfugkhkckgijljhgxjgjgffhgkgjxhxjgjcjckvjgghkhkgjgjfhfhfhffrusufsflslrsyfhldufñlñudtoqdyhjjxkgsgjfktwlyfñujxjxhlxlhdktstedoyñfuyñflldytidoyeyljjcñcjluffñui5woyepurñfuñufldyrajuñdlydstdyñudñydktshñxjcñydiw5uñfitwoyeoyeñufñfuñifjñufhlsyñeifññydoysitaiwtuñdyñdlsyltslsyoyeylsuñdñjjcyldlyslatlysñudidñjdñfjñjjxlhsmzhmzjjdjdlhdñhjdñjdñjddñhflhuñfhxltkds4urayraylraluarularuñstuñtsuñtsultsuñtsuñstñitsñktssistustlulsrustlularyralultalutslutajltsñgskjlgzljg?g o uguhxputxipyfugxiñhxiñhfuñdguldthgksjmgdjmgkhdjlgdjlgd

pduoyditsyafylrayoraourauptautospustistiptsñitsñitsñitsiptsiteitdustuñtsuñtsñitwiñstñitwñitsñstuuñrsoursurosoustjlsrlutejlgsjlstjfsjlgsultsjgzjñgsññkdylfhkñdgjlfshkadmjgsuñstñkydñkydñiykdhiñstñitsuñtsisñtñtieñietñietñiteñiwtñitskñgsiñteuñwrkñsturaluglsuñtwjlfalfjalhadoyfutdllgdñitswtkgsñktjrajtsurwñwñutiñtsiwñtuwñturqlñitwualtayoryarluarlietite

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Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
just olya [345]

Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Alex787 [66]

Answer:

a) 1.092 m/s

b) 0.33 m

c) 0.25 m

Explanation:

To start with, from the formula of wave, we know that

v = f λ, where

v = velocity of wave

f = frequency of the wave

λ = wavelength of the wave

Again, on another hand, we know that

T = 1/f, where T = period of the wave

From the question, we are given that

t = 2.7 s

d = 0.66 m

λ = 5.9 m

Period, T = 2 * t

Period, T = 2 * 2.7

Period, T = 5.4 s

If T = 1/f, then f = 1/T, thus

Frequency, f = 1/5.4

Frequency, f = 0.185 hz

Remember, v = f λ

v = 0.185 * 5.9

v = 1.092 m/s

Amplitude, A = d/2

Amplitude, A = 0.66/2

Amplitude, A = 0.33 m

If the other distance travelled by the boat is 0.5, then Amplitude is

A = 0.5/2

A = 0.25 m

6 0
3 years ago
Objects the exhibit projectile motion follow a ____________ path.
s344n2d4d5 [400]
B. Parabolic

This is because an object that is in projectile motion has gravity acting on it.
6 0
3 years ago
The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by the work of which of the following scientists?
borishaifa [10]

Answer

Correct Answer is B(Edwin Hubble)

Explantion

The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by Edwin Hubble because both have worked in the feild of astrophysics.

Subrahmanyan Chandrasekhar was an astrophysicist. He belons to India. He is known for his work on the theoretical structure and evolution of stars. Subrahmanyan Chandrasekhar with William Fowler won the Nobel Prize in Physics in 1983 largely for this early work. He has worked in many other areas within theoretical physics and astrophysics.

Edwin Hubble was an astronomer who belongs to America. He is known for Hubble Law. Hubble discovered that many objects which was thought to be clouds of dust and gas  were actually galaxies beyond the Milky Way. He has an important contribution in establishing the fields of observational cosmology and extra galactic astronomy


3 0
3 years ago
Read 2 more answers
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