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LenKa [72]
3 years ago
5

Average speed of a runner runs 500 .o min. 1.6 min is

Physics
1 answer:
ch4aika [34]3 years ago
3 0
<span>I believe your answer will be 5.2 meters </span>
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At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
What can the magnetic field be used for​
deff fn [24]

Answer:

the magnetic field can be used to make electricity

Explanation:

Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy

8 0
3 years ago
What is the Validity measures of a basketball bouncing
kykrilka [37]

This study was aimed at testing the construct validity of the basketball basic motion skills test instrument (ITK GDBB). The research used descriptive method of 3 basketball experts in the city of Cimahi; 3 experts are the expert in basketball. The instrument used was the ITB GDBB developed by Silvy (2019) consisting of top passing, bottom passing, top service, bottom service, chest passing, bounding passing, overhead passing, and leading ball (dribbling). This instrument consists of 76 items that cover 4 domains in basketball, namely chest pass, overhead pass, bound pass, and dribbling. The validity method used the construct validity of different power types. For the reliability method, it used the Kuder Ricardson (KR) and Objectivity analysis. The results of the construct validity analysis of a total of 76 items show that the score is ranged from 0.67 to 1.00. The construct validity value of 71 items in the basketball game is in the high category (= 1.00), 5 items are in the sufficient category, the relativity score is ranged from 0.75 to 0.98, and the objectivity score is ranged from 0.89 to 0.95. The conclusion is that this test instrument can be used as a standardized basic motion skill test for standardized large ball games for validity in basic motion skills in basketball games for grade VII junior high school students.

3 0
3 years ago
Europe and North America are moving apart by about 5 m per century. As the continents separate, new ocean floor is created along
Alinara [238K]

Answer:

Area = 25km² in a century.

Explanation:

Given

Spread Rate = 5m/century

Length of rift = 5000km

Convert to metres

Length of rift = 5000 * 1000m

Length of rift = 5,000,000m

In one century, the additional area to Atlantis = (Rift Length) * (Spread rate) * 1 century

Atea = 5,000,000 m * 5m/century * 1 century

Area = 25,000,000m² in a century

----- Convert to km²

Area = 25,000,000km² * 1m²/1,000,000km²

Area = 25km² in a century.

Hence, the total area of new ocean floor created in the Atlantic each century is 25km²

6 0
3 years ago
Why do we need to learn about median and mode?
siniylev [52]
We need to learn about median and mode because they are both important in statistical data and statistical data is what shows us important information about the world around us.
3 0
3 years ago
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