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3 years ago

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf. t/f

2 answers:

7 0

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf is false. The kinetic energy on the shelf is zero because it is not in action.

algol [13]3 years ago

6 0

Answer: The given statement is false.

Explanation:

Kinetic energy is defined as the energy obtained by an object because of its motion.

For example, moving a pulley on the road.

Whereas energy of an object obtained due to its position is known as potential energy.

For example, lifting a book from table to the shelf shows position of the book has changed.

Therefore, we can conclude that the statement kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf, is false.

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4 The temperature at the center of the sun is 1.55 x 10' K. Express this in degrees

Sloan [31]

Answer:

The answer is D

Explanation:

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4 years ago

Tell whether the following statements are true or false:

Digiron [165]

Answer:

1)true

2)false

3)true

4)true

5)false,cz amatter is made of atoms while light is an electromagnetic radiation.

8 0

3 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin

Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

w₀B = ½ w₀

αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

t´ = t - t_1

et's write the equation of particle B

θ = θ₀ + w₀B t´ + ½ αB t´2

replace

θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0

3 years ago

An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati

Viefleur [7K]

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

8 0

3 years ago