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3 years ago

What is the kinetic energy of a 150 kg bear running at 3 m/s?

Physics

1 answer:

OverLord2011 [107]3 years ago

5 0

Explanation:

KE = ½ mv²

KE = ½ (150 kg) (3 m/s)²

KE = 675 J

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How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory

xeze [42]

Question:

1) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

2) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

3) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

4) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

Answer:

The correct option is;

3) The Universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Explanation:

With the temperature measurement carried out using the CSIRO radio telescope, Astronomers have been able to determine a temperature difference in the universe from 5.08 Kelvin 7.2 billion light years away to 2.73 Kelvin in the Universe today, which is in support of the Big Bang theory that as the Universe expanded from a state of extreme temperature that cools down as the Universe expands or the cosmos disperses.

8 0

2 years ago

Which is NOT true of the Big Bang Theory?

grigory [225]

Answer:

1. it explains what's happening im the universe now

3 0

2 years ago

Across:

vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

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6 0

2 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

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)