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1 year ago

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Consider (a) an electron (b) a photon, and (c) a proton, all moving in vacuum. Choose all correct answers for each question. (iv

) Which carry momentum?

1 answer:

DiKsa [7]1 year ago

4 0

Electron, photon and proton carry momentum .

De Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.
De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ ...(1)
Momentum is the product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

Equation (1) can be written as $\lambda=\frac{h}{p}$ where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant

For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

From equation (1) and (2) we can conclude that electron, photon and proton carry momentum .

Learn about more photon momentum here :

brainly.com/question/13112527

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

Which of the following is not a colligative property? a. osmotic pressure b. lattice energy c. vapor-pressure lowering d. boilin

Gre4nikov [31]

Answer:

b) lattice energy

Explanation:

A solution is said to have colligative property when the property depends on the solute present in the solution.

Colligative property depend upon on the solute particle or the ion concentration not on the identity of solute.

osmotic pressure, vapor pressure lowering , boiling point elevation and freezing point lowering all depend upon solute concentration so they will not have colligative property so, the answer remains option 'b' which is lattice energy.

8 0

3 years ago

A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an

Vanyuwa [196]

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

6 0

3 years ago

What are the two types of physical fitness?<br>

Marina86 [1]

<em><u>your </u></em><em><u>question</u></em><em><u>:</u></em><em><u> </u></em>

<em>What are the two types of physical fitness?</em>

<em><u>answer:</u></em>

<em>The</em><em> </em><em>two </em><em>types </em><em>of </em><em>phys</em><em>ical </em><em>fitn</em><em>e</em><em>s</em><em>s</em><em> </em><em>are </em><em>Health-related</em><em> physical fitness and Performance-related physical fitness.</em>

4 0

3 years ago

Suppose that you have a rod that is positively charged, a rod that is negatively charged and a stand that allows rotation. If yo

amm1812

Answer:

Explanation:

A charge is produced when an atom losses or gains an electron. The law of static electricity states that like charges repels, while unlike charges attracts.

1. To determine the charge on the polystyrene rod.

Place the polystyrene rod on the non-conducting rotating stand, and bring the positively charged rod close to it. If attraction occurs, it shows that it is oppositely charged. If repulsion occurs, it shows that it is positively charged.

Bringing a negatively charged rod close to the rotating polystyrene rod would attract it if the charge is opposite. But if the charge on the two rods are the same, repulsion occurs.

2a. When the polystyrene rod is positively charged, it would attract the negatively charged rod but repel the positively charged rod.

b. When the polystyrene rod is negatively charged, it would repel the negatively charged rod but attract the positively charged rod.

c. When the polystyrene rod is uncharged, no reaction would be observed when either the positively charged or negatively charged rod is brought close to it.

3 0

3 years ago