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3 years ago

9

How does speed relate to motion energy?

2 answers:

sleet_krkn [62]3 years ago

8 0

The following equation is used to represent the kinetic energy (KE) of an object.

Brainliest?

Murljashka [212]3 years ago

4 0

the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

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Problem 9.49: Air enters the turbine of a gas turbine at 1200 kPa, 1200 K, and expands to 100 kPa in two stages. Between the sta

Nitella [24]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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3 years ago

Suppose a large housefly 3.0 m away from you makes sound with an intensity level of 40.0 dB. What would be the sound intensity l

melisa1 [442]

Answer:

Explanation:

Let the intensity of the noise be represented by I

Given that

40dB = 10 log 10 ⁡ ( I /I•) ........ 1

I• is the lowest or threshold intensity of sound made.

I represents the intensity of the sound/ noise

The intensity of noise of 1000flies will be

β = 10 log 10 ⁡(1000I/I•)

Open up the bracket

β = 10 log 10(1000)+ 10 log 10(I/I•)

10 log 10(10^3)+10 log 10(I/I•)

3×10(10 log 10) +10 log 10(I/I•)

Recall, 10 log 10 = 1

30×1 + 10 log 10(I/I•).........2

Put equation 1 into 2

β =30+40

= 70db

5 0

3 years ago

What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?

madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

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2 years ago

An amusement park ride consists of a rotating circular platform 10.9 m in diameter from which 10 kg seats are suspended at the e

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3 years ago

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, $F_c=\frac{mv^2}{R}$

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

$f=\mu N$

As there is no vertical motion, therefore, $N=mg$ . So,

$f=\mu mg$

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816$

Therefore, option (c) is correct.

7 0

3 years ago