PHYSICS HIGH SCHOOL

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.9 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 70.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 40 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answers

Answer 1

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

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Explain the energy transformations that occur when breaking in a hybrid
vehicle.

Answers

it occur kinetic energy

when breaking in a hybrid

vehicle.

hope it helps

Answer:

it occur kinetic energy when breaking in a hybrid

vehicle .

hope it is helpful to you ☺️

What's the value of 57,281 joules in Btu

Answers

54.3 BTU is what I got

54.291912

I hope this helps :)

Frequency refers toa. how much sweat is lost during exercise.
b. the type of exercise a person is doing.
c. the intensity of the workout.
d. how many times per week a person exercises.

Answers

Frequency refers to how many times per week a person exercises. The correct option among all the options that are given in the question is the fourth option or option "d". Frequency is actually the rate at which the person does the work or rather the average. I hope this answer has come to your help.

Light traveling through air at 3.00 · 10^8 m/s reaches an unknown medium and slows down to 2.00 · 10^8 m/s. What is the index of refraction of that medium? n =

Answers

v 1= 3.00 · 10^8 m/s
v 2 = 2.00 · 10^8 m/s
The index of refraction:
n = v 1 / v 2 = 3.00 · 10^8 m/s : 2.00 · 10^8 m/s = 1.5
Answer:
The index of refraction of that medium is 1.5

Answer

The index of refraction of medium is 1.5 .

Explanation:

Equation for index of refraction is given by .

$n = (c)/(v)$

Where n is index of refraction in medium , v is speed of light in the medium and c is the speed of light in air .

As given

Light traveling through air at $3.00* 10^8$ m/s reaches an unknown medium and slows down to $3.00* 10^8$ m/s.

$c = 3.00* 10^(8)$

$v = 2.00* 10^(8)$

Putting the values in the formula

$n = (3.00* 10^(8))/(2.00* 10^(8))$

Solving the above

n = 1.5

Therefore the index of refraction of medium is 1.5 .

Millikan measured the electron's charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.60 × 10-19 C. Find the charge on an oil drop weighing 1.00 × 10-14 N and suspended in a downward field of magnitude 2.08 × 104 N/C.

Answers

Answer:

$4.8\cdot 10^(-19) C$

Explanation:

For a drop in equilibrium, the weight is equal to the electric force (in magnitude):

$W = F_e$

where here we have

$W=1.00\cdot 10^(-14)N$ is the weight of the drop

$F_e$ is the magnitude of the electric force, which can be rewritten as

$F_e = qE$

where

q is the charge of the oil drop

$E=2.08 \cdot 10^4 N/C$ is the magnitude of the electric field

Substituting into the equation and solving for q, we find the charge of the oil drop:

$q=(W)/(F_e)=(1.00\cdot 10^(-14)N)/(2.08\cdot 10^4 N/C)=4.8\cdot 10^(-19) C$

How can researching travel time, routes and weather conditions benefit drivers who are planning a road trip?

Answers

Explanation:

They will be alerted of the traffic conditions. They can choose the route according to the weather condition. They will be aware if there is any road construction going on or accident happened. These awareness will save their time and avoid delays. their journey can be more comfortable and less trouble some.

Explanation:

Researching these options time, routes and weather conditions and even more gives the driver ample information about the journey, and places the driver in better position for decision making, with this information and data afore hand he can decides the best route to take should there be any gridlock, he even decides where or not to embark on the journey based on the prevailing weather conditions,

Regarding time of the journey he is well disposed for proper planning and scheduling.

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