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3 years ago

The ability to move or change an object or what a wave carries

Physics

1 answer:

Sever21 [200]3 years ago

5 0

Answer: The ability to move or change an object or what a wave carries is called Energy

Explanation: Waves are disturbances in physical quantities. Example of waves are light waves, sound waves, or transverse oscillations of a string. These disturbances use energy to create and propagate, for it to move the constituent particles or change the electric or magnetic fields. Therefore, power of a wave is therefore, energy transported divided by unit time caused by the oscillations of a particular wave. The derivation of a formula for the power depends on the medium -- for light waves, the power is measured by the pointing vector, whereas for oscillations on a string, the power can be computed directly by balancing forces through the application of newton law. However, for all types of waves, the formula and physical meaning of the power takes similar forms, typically depending on the square amplitude of the waves among other factors.

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Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False

kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

$\vec{L} = I \vec{\omega}$

where $I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2$ is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

$\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2}$ so $\vec{\omega} \neq 0$ unless $\vec{r}$ ║ $\vec{v}$ .

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

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2 years ago

Explain why nuclear fission and nuclear fusion release large amounts of energy

levacccp [35]

Answer:

Because of the formula $E=mc^2$

Explanation:

In this problem we are describing two different processes:

Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

$E=\Delta mc^2$

where:

E is the energy released in the reaction

$\Delta m$ is the mass defect, the difference between the final total mass and the initial total mass

$c=3.0 \cdot 10^8 m/s$ is the speed of light

From the formula, we see that the factor $c^2$ is a very large number, therefore even if the mass defect $\Delta m$ is very small, nuclear fusion and nuclear fission release huge amounts of energy.

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2 years ago

A 2.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from rest and lands on the cart. The

anzhelika [568]

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My nickname - Lovely

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3 years ago

How can you increase the gravitational potential energy (GPE) of an empty shoe box on the bottom shelf of a bookcase

Anettt [7]

Answer:

Add items to the box

Explanation: I did the test

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2 years ago

Read 2 more answers

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

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3 years ago