1. Starting today. I will now change my behavior/s like ______________ because ________________

A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite

DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x 4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π

≥ . 5254 x ⁻²⁵

h / λ ≥ . 5254 x ⁻²⁵

6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ

12.56 x 10⁻⁹ ≥ λ

longest wave length = 12.56 n m

6 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi

max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

3 0

3 years ago

When you are sitting a few feet from a fire, your skin feels warm. What forms of heat transfer are acting to transfer heat from

algol [13]

<span>In the question,' when you are sitting a few feet from the fire, your skin feels warmed. What form of heat transfer are acting to transfer heat from the fire to your skin, the correct option is A, that is, convection and radiation. Heat transfer is defined as the exchange of thermal energy between physical systems. The rate at which the heat is transfer depends on the temprature of the system and the properties of the intervening medium through which the heat is been transfered. There are three basic modes of heat transfer, these are: conduction, convection and radiation. Conduction is defined as the transfer of heat between two bodies through physical contact. When two bodies which have different temprature come in contact, there will be a transfer of heat energy between them until the two of them have the same temprature. Conduction usually occurs in solids and liquids; it occurs in gases also but it is extremely slow. Convection is the process by which heat is transfer in fluids, that is, liquids and gases. This is how convection operates: when a fluid is heated, it expands and it becomes lighter, this makes it to rise upward and move to the cooler part of the container, as it rises, it will be replaced by the unheated surrounding particles. This cycle continues until heat is evenly distributed all through the fluid. There are two types of convection: natural and forced convection. The heating of the earth surface by the sun ray is an example of natural convection while the air conditioner we use at home operates by mean of forced convection. Both conduction and convection require matter for heat transfer. Radiation is the transfer of heat from one place to another through electromagnetic waves. The hot body transfer heat by emitting electromagnetic waves. The properties of the electromagnetic waves depend on the temperature of the body. The higher the temperature the more intense the rate of emission of radiation. Radiation can occur in all objects and does not require matter for heat transfer. The heat of the sun reaches the earth surface by means of radiation. In the question given, as the air surrounding the fire were heated they rise and were replaced by the unheated air particles. The continuation of this cycle makes the heat energy to be transferred to the objects around. Thus, the heat from the fire was transferred via convection and radiation. </span>

8 0

3 years ago

Read 2 more answers

What part of the plant takes in carbon dioxide?

murzikaleks [220]

The answer is number 2 stomata.

4 0

3 years ago

Read 2 more answers

1. Starting today. I will now change my behavior/s like ______________ because _________________

1. Starting today. I will now change my behavior/s like because ___