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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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34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf
ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

d_1 = \frac{1}{2}(0.6) t^2

also the distance moved by passenger

d_2 = \frac{1}{2}(1.2) t^2

so we will have

d_1 + 5 = d_2

0.3 t^2 + 5 = 0.6 t^2

0.3 t^2 = 5

t = 4.08 s

3 0
3 years ago
Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e
ANEK [815]

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

\dfrac{K-273}{100}=\dfrac{F-32}{180}

Now by putting the values

\dfrac{K-273}{100}=\dfrac{F-32}{180}

\dfrac{0-273}{100}=\dfrac{F-32}{180}

So F= - 459.67°F

So we can say that 0 K is equal to  - 459.67°F.

So the our option e is correct.

3 0
4 years ago
Near which location in New York State would a geologist have the greatest chance of finding dinosaur footprints in the surface b
Likurg_2 [28]

Answer:

1

Explanation:

41° 10' N latitude, 74° W longitude

5 0
3 years ago
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0
3 years ago
I tried but i don’t understand! please help asap!!!
Semmy [17]

Answer:

~218 electrons

Explanation:

(-3.5 * 10^-17 C)/(-1.602 * 10^-19 C) = 3.5/1.602x10^-2 = 218.47...

5 0
3 years ago
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