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1 year ago

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A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 32 m at t2 = 2.0 s .

Answer the following by considering ratios, without computing the car's velocity. 1. What is the car's position at t = 1.0 s ? 2. What will be its position at t = ? answers pleaseeeee

1 answer:

Lerok [7]1 year ago

3 0

Without computing the car's velocity, the car's position at t = 1.0 s will be x = 16m

<h3>Position - Time Graph</h3>

The gradient of the position time graph is speed.

Given that a car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 32 m at t2 = 2.0 s.

The gradient m = Δx ÷ Δt

m = (32 - 0) / (2 - 0)

m = 32 / 2

m = 16 m/s

Without computing the car's velocity, the car's position at t = 1.0 s will be

32 / 2 = x / 1

x = 16 m

Therefore, without computing the car's velocity, the car's position at t = 1.0 s will be x = 16m

Learn more about Position Time Graph here: brainly.com/question/18388330

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On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

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Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

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Which statements about velocity are true?

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<span>the speed of a direction</span>

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3 years ago

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After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th

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Answer:
These are the two statements with scientific facts that explain the described phenomenon
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Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.

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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.

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3 years ago

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