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3 years ago

John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

Physics

1 answer:

sdas [7]3 years ago

6 0

Average speed = distance traveled / time

average speed = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min

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eduard

Answer:

3. A

4. C

Explanation:

4 0

3 years ago

A jet makes a landing traveling due east with a speed of 120 m/s .

vesna_86 [32]

Average acceleration over a time interval lasting $\Delta t$ is

$a_{\rm ave}=\dfrac{\Delta v}{\Delta t}$

where $\Delta v$ is the difference in the jet's final and initial velocities. It's coming to a rest, so

$a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}$

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).

7 0

4 years ago

A centrifuge rotor (hollow disk) is rotating at 10,000 rpm is shut off and brought to rest by a constant frictional torque of 1.

Y_Kistochka [10]

Answer:

The no of revolutions rotor turn before coming to rest is 1,601.1943 and time taken is equal to 19.21 seconds

Explanation:

here we know that torque = I×α

α= angular acceleration

I = moment of inertia of hollow disc = m× $k^{2}$

given that m=4.37kg

k=0.0710m

torque=1.2Nm

$w_{o}=10000\times \frac{\pi}{30} rad /s$

$\alpha =\frac{torque}{I}$

from the above equation we can calculate the angular acceleration of the hollow disc .

since $w^{2}-w_o^{2} =2\alpha$ $\theta$

from this above equation $\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }$

no of revolutions = $\frac{\theta}{2\pi}$ = 1,601.1943.

Now to calculate time we know that time = $\frac{2\theta}{w+w_{o}}$

so upon calculating we will be getting t=19.21 seconds

5 0

3 years ago

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

Student one used bowling ball A in a bowling game against Student 2, who used bowling ball B. Use Newton’s Three Laws of Motion

Jlenok [28]

Answer:

Newton’s Three Laws of Motion has a great impact.

Explanation:

Newton’s Three Laws of Motion has a great impact on the bowling game for the 2 students. When the student one throw ball to the student 2, the ball decrease its speed due to the gravity and opposing air. If these forces are removed from the system the ball will continue its motion till another force is applied on it. When the force applied to the ball it produces acceleration in the direction to the applied force. If the ball touches the ground it bounce back with equal force which is a reaction of the ground.

8 0

3 years ago