All categories

3 years ago

Which statement is true?

Physics

1 answer:

marshall27 [118]3 years ago

3 0

Answer:

B.The galaxies are moving away from us.

Explanation:

Universe is expanding from the time of Big bank.

AS the universe expand Galaxies get away from each other.Latest studies stay that the acceleration of universe is uncreasing due to Dark Energy.

You might be interested in

What is unusual about the results of mass determinations of clusters of galaxies?

Art [367]

Answer:

I think it's bigger than most galaxies

3 0

2 years ago

Read 2 more answers

A pump is used to transport water to a higher reservoir. if the water temperature is 15ºc, determine the lowest pressure that ca

Reika [66]

Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation.

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at $15^{\circ}$ is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.

7 0

3 years ago

An unknown substance has a mass of 15 g and a volume of 4 cm3. Calculate the density of the unknown substance.

myrzilka [38]

Answer:

Density is 3.75g/cm3

Explanation:

Density = mass ÷ volume

= 15g ÷ 4cm3

= 3.75g/cm3

5 0

3 years ago

A quantitative description of kinematics involves using ___ th describe the motion

Semmy [17]

Answer:

numbers

Explanation:

Virtually all unimaginable processes can be described as the movement of certain objects. To analyze and predict the nature of the movements that result from the different kinds of interactions, some important concepts such as momentum, force and energy have been invented. If momentum, force, and energy are known and expressed in a quantitative way (that is, by numbers) it is possible to establish rules by which the resulting movements can be predicted.

4 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:

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