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CaHeK987
1 month ago
10

Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from

the center of the circular platform as Isaac, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed?
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0
The question lacks complete details or specifications. Here is the missing information: 1. impossible to establish 2. half of Isaac's 3. identical to Isaac's 4. double Isaac's The angular velocity of Feng will match that of Isaac. Thus, the correct choice is option 3.
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You're riding a unicorn at 25 m/s and come to a uniform stop at a red light 20m away. What's your acceleration?
Ostrovityanka [3204]

The result is -15.625 m/s².


Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:


a = \dfrac{vf-vi}{t}

Where:

vf = final velocity

vi = initial velocity

t = time

Let’s examine the information provided in your query:

Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.


However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.


The time can be calculated using the kinematics equation:

d = \dfrac{vi+vf}{2} *t


We derive the equation by substituting the known values first.

20m = \dfrac{25m/s+0m/s}{2}(t)

20m = 12.5m/s{2}(t)

\dfrac{20m}{12.5m/s}=t
1.6s=t

The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.


a = \dfrac{0m/s-25m/s}{1.6s}

a = \dfrac{-25m/s}{1.6s}

a = -15.625m/s^{2}


It is important to note that the acceleration is negative, indicating the vehicle slowed down.

8 0
1 month ago
Read 2 more answers
A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha
serg [3582]

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}

8 0
1 month ago
A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,
Sav [3153]
Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.
6 0
1 month ago
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
Yuliya22 [3333]

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x\frac{1}{3}  x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

=  2π x  11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x  2π x  11 / 60 rad / s

= 128.23 x 10⁵  kgm² s⁻¹.

4 0
2 months ago
The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring
Sav [3153]

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

E=PE_{2}-PE_{1}

E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2

Substitute the values into the formula

E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2

E=0.0072\ J

E=7.2\ mJ

Therefore, a total of 7.2 mJ is needed to extend the pen.

7 0
2 months ago
Read 2 more answers
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