Y=k/x, x is halved.<br> what happens to the value of y

A second-grade teacher dropped a box of paper clips and they scattered all over the floor. She then asked her students, "Why wil

Yuliya22 [10]

The magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips.

<h3>Attraction of magnets</h3>

The like poles of magnets repel while unlike poles of magnets attracts. Magnets attracts irons or metallic materials.

The paper clips are mettalic or made of iron and hence the magnet will attract them.

There we can conclude that the magnet will be a useful tool to pick the paper clips because the magnet can attract the paper clips and will help to gather them together for easy picking.

Learn more about magnets here: brainly.com/question/14997726

8 0

2 years ago

Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per

Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 $^{\circ}$ to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

Which of the following is the best reason that trees care important for overall air quality?

Marina86 [1]

Trees are important because oxygen

4 0

3 years ago

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0

3 years ago

Y=k/x, x is halved. what happens to the value of y