Q. A car is traveling

An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

(B) 
(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev 

(C) 
(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s

3 0

3 years ago

Read 2 more answers

8. Noticing how much food is on your lunch tray is a quantitative observation because

zhenek [66]

Answer:

you are making an observation that uses numbers.

Explanation:

5 0

3 years ago

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint conce

WINSTONCH [101]

Answer:

a) x₁ = 290.50 feet , x₂ = 169.74 feet , b) v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis fr = m a

Y Axis N-W = 0

N = W = mg

The force of friction has the expression

fr = μ N

We replace

μ mg = ma

a = μ g

g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ a (feet / s2)

0.599 19.168

0.536 17,152

0.480 15.360

0.350 11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

v² = v₀² - 2 a x

When the speed stops it is zero

x₁ = v₀² / 2 a₁

Let's reduce speed

v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

x₁ = 80.667² / (2 11.2)

x₁ = 290.50 feet

The minimum braking distance

x₂ = 80.667² / (2 19.168)

x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

0 = v₀² - 2 a x

v₀ = √2 a x

We calculate the speed for the two accelerations

v₀₁ = √ (2 11.2 155)

v₀₁ = 58.92 feet / s

v₀₂ = √ (2 19.168 155)

v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

5 0

3 years ago

Massive stars terminate in a brilliant explosion called a

BigorU [14]

super nova is named for such explosion

6 0

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:

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