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madreJ [45]
3 years ago
12

A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

rcular motion is a constant 50.00 cm, and the time it takes the stopper to make one full revolution is 0.2500 seconds, what is the tangential speed of the stopper?
Physics
1 answer:
alexandr402 [8]3 years ago
3 0

Answer:

v = 12.57 m/s

Explanation:

As we know that the radius of the circular motion is given as

R = 50.00 cm

time period of the motion is given as

T = 0.2500 s

now we know that it is moving with uniform speed

so it is given as

v = \frac{2\pi R}{T}

now plug in all data

v = \frac{2\pi(0.50)}{0.25}

v = 12.57 m/s

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When vibrational motion in an object increases, which is a true statement?
ss7ja [257]
Hello friend!!

We know that kinetic energy is the energy possessed due to the motion of the object. And we know if the object is in a fast motion then the temperature would be high, whereas if the object is slow in motion then it will have lower temperature. So we know that the kinetic energy is indirectly related to temperature.From our knowledge we can conclude that HIGHER THE TEMPERATURE, HIGHER THE KINETIC ENERGY and LOWER THE TEMPERATURE, LOWER THE KINETIC ENERGY.
Hence, the answer to your question here is,a.kinetic energy, temperature, and thermal energy increase. 
Hope it helps!!All the best!!
5 0
3 years ago
1. The geologic time scale divides time into years and centuries, true or false
hodyreva [135]

Answer:

The answer is "False"

Explanation:

The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.

Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.

5 0
2 years ago
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
ratelena [41]

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

4 0
3 years ago
Please help on this one?
Schach [20]

Chemical energy. Elimination is useful to answer this question.

6 0
2 years ago
Which object has more gravitational potential energy
Sonja [21]

Answer:

need more context

Explanation:

7 0
3 years ago
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