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3 years ago

20 POINTS!!!

Physics

1 answer:

bogdanovich [222]3 years ago

5 0

a) Average velocity for her run north: 2.68 m/s north

b) The average velocity for the whole trip is zero

c) The average speed for the whole trip is 1.95 m/s

Explanation:

In this part of the problem we only want to consider the part when Micha is running north.

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the displacement (a vector connecting the initial position to the final position of motion)

t is the time

For the part running north, we have

$d=2 mil \cdot 1609 = 3218 m$ north is the displacement

$t = 20 min \cdot 60 = 1200 s$ is the time interval

Substituting,

$v=\frac{3218}{1200}=2.68 m/s$ north (we have to specify also the direction, since velocity is a vector)

As we said, average velocity is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time

If we consider the whole trip, however, the displacement is zero:

d = 0

Because Micha returns home, so the initial position corresponds to the final position of motion. Therefore, the average velocity is also zero:

v = 0

The average speed is given by

$s=\frac{d}{t}$

where

d is the distance covered (the total length of the path, regardless of the direction)

t is the time interval

Since MIcha ran 2 miles north and then 2 miles back, the total distance is

$d=2 mi + 2mi = 4 mi \cdot 1609 = 6436 m$

And the time taken is

$t=20 min + 15 min + 20 min = 55 min \cdot 60 = 3300 s$

So, the average speed is

$s=\frac{6436}{3300}=1.95 m/s$

And since speed is a scalar, there is no need to specify a direction.

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

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3 years ago

How many centimeters are there in meter? b. 10 c. d 1000 e. 10000 100 2. A centimeter is equal to 1 inch b.½inch C 1/2.54 inch

astra-53 [7]

Answer:

a) There are 100 centimeters in 1 meter.

b) $\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

Explanation:

a) We have the conversion

1 m = 100 cm

So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

$1cm=\frac{1}{2.54}inch$

$\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

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3 years ago

A jet aircraft is traveling at 262 m/s in hor-

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Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

= 4862635.79 + 3610.32

= $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$ .

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3 years ago

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:

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