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Softa [21]
2 years ago
12

The angle turned through the flywheel of a generator during a time interval t is given by theta = at + bt^3 -ct^4, where a, b, a

nd c are constants. What is the expression for its angular acceleration?
Physics
1 answer:
Alenkinab [10]2 years ago
5 0

Answer:

α = 6bt - 12ct^2

Explanation:

The angle, θ, has been given as:

θ = at + bt^3 -ct^4

To obtain angular acceleration, α, we have to differentiate the angle twice, with respect to time, t.

Differentiating once will yield the angular velocity, ω:

ω = dθ/dt = a + 3bt^2 - 4ct^3

This can then be differentiated to obtain angular acceleration:

α =  dω/dt = 6bt - 12ct^2

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(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
Anastaziya [24]
3 is jxhuneuxndnzixbf
7 0
2 years ago
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A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
2 years ago
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.
lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\

Δtm is  time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the  sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0
2 years ago
When a car goes around a circular curve on a horizontal road at constant speed, what force causes it to follow the circular path
Reil [10]
Centripetal force allows cars to follow a circular path.
4 0
2 years ago
What is the de broglie wavelength of a 149-g baseball traveling at 95.4 mph? (1 mile = 1.609 km, h = 6.63 × 10–34 j·s)?
boyakko [2]

The solution for the problem is:

Wavelength = Planck’s constant/(mass*velocity) 

Planck’s constant= 6.63*10^-34 with units of J-s or kg-m^2/s^2-s 

mass = 149g = 0.149 kg 

velocity = 95.4.mi/1hr(1609.3m/1mi)(1hr/3600sec) = 42.65m/s 

h/mv = 6.63*10^-34 kg-m^2/s^2-s/(42.65m/s*0.149kg)

wavelength = 1.04 *10^-34 m

5 0
2 years ago
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