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3 years ago

Calculate the distance in km that Charlie runs if he maintains the average speed from question 2 for 1 hour

Physics

2 answers:

egoroff_w [7]3 years ago

5 0

Sorry, but what is question 2?

weqwewe [10]3 years ago

4 0

Distance covered by charlie= 2 km

Explanation:

we know that

$d= v\times t$

d= distance

t= time= 1 hour

v= velocity= 2 km/h

so distance d= 2 km/h (1 h)

d=2 km

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Explanation:

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2 years ago

Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?

musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

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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:

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

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

Why do you think a human needs only one egg at a time

babunello [35]

Because we can reproduce with just one and if more eggs were released, the female reproductive life (which is 40 years) would be shorter and so there would be less time for every female human to reproduce since the amount of eggs is limited. (Also fraternal twins come from 2 eggs released at the same time)

7 0

3 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ