A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse

(a) The de Broglie wavelength is approximately 5.175 × 10⁻¹⁵  meters. The wavelength is lesser than the distance of closest approach

(b) It is proper to treat the alpha particle as a particle and not as wave because the distance of closest approach is much larger than and not comparable to its wavelength for the alpha particle for the alpha particle to be treated as a wave

The given parameters are;

The kinetic energy of the alpha particles = 7.70 MeV = 1.23368 × 10⁻¹² J

The distance from the gold nucleus the alpha particles reach = 29.5 fm

(a) The de Broglie wavelength of a particle is given as follows;

[tex]\mathbf{\lambda = \dfrac{h}{p}}[/tex]

Where;

λ = The wavelength

h = Planck's constant = 6.62607004 × 10⁻³⁴ m²·kg/s

p = The momentum of the particle = Mass of an electron, m × Velocity, v

The mass of an alpha particle, m ≈ 6.645 × 10⁻²⁷ kg

Therefore;

[tex]\lambda = \dfrac{h}{m \times v}[/tex]

The kinetic energy of the alpha particle, K.E. = (1/2)·m·v²

∴ v = √(2 × K.E./m)

Therefore;

[tex]\lambda = \dfrac{h}{m \times \sqrt{2 \times \dfrac{K.E.}{m} } } = \dfrac{h}{ \sqrt{2 \times m \times K.E.} }[/tex]

Plugging in the values of the variables gives;

[tex]\lambda = \dfrac{6.62607004 \times 10 ^{-34} }{ \sqrt{2 \times 6.645 \times 10 ^{-27} \times 1.23368 \times 10^{-12} } } \approx 5.175 \times 10^{-15}[/tex]

The de Broglie wavelength of the alpha particle, λ ≈ 5.175 × 10⁻¹⁵ m

The distance of closest approach = 29.5 fm = 29.5 × 10⁻¹⁵ m

Compared to the distance of closest approach, the wavelength of the alpha particle is lesser than the distance of closest approach

(b)  Given that the distance of closest approach is six times larger than the wavelength of the alpha particle, and alpha particle behaving as waves are expected to approach closer to the gold nucleus in the region of their wavelength before deflection, therefore, the larger distance of closest approach is indicative of a charged particle to charged particle interaction, and therefore, particle behavior of alpha particles.

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I think it is a education tips

Answer: The mass of copper liberated is 0.196 g.

Explanation:

The oxidation half-reaction of copper follows:

[tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]

Calculating the theoretical mass deposited by using Faraday's law, which is:

[tex]m=\frac{M\times I\times t(s)}{n\times F}[/tex] ......(1)

where,

m = actual mass deposited = ? g

M = molar mass of metal = 63 g/mol

I = average current = 2 A

t = time period in seconds = 5 min = 300 s (Conversion factor: 1 min = 60 sec)

n = number of electrons exchanged = 2

F = Faraday's constant = 96500 C/mol

Putting values in equation 1, we get:

[tex]m=\frac{63 g/mol\times 2A\times 300s}{2\times 96500 C/mol}\\\\m=0.196g[/tex]

Hence, the mass of copper liberated is 0.196 g.

Answer:

in order to know the length of something

Answer:

Slowly and smoothly lol

Answer:

S.I. on Rs. 1600 = T.D. on Rs. 1680. Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%. Time =100 x 80year=1year = 4 months.1600 x 153

Explanation:

Is this you are?

Answer:

Efficiency of a machine is defined as the ratio of output work to input work in a machine . It is expressed in percentage and denoted by

η ( eta).

Answer:

Distancia, D = 170,000 metros

Explanation:

Dados los siguientes datos;

Para encontrar la distancia recorrida, en metros;

Matemáticamente, la distancia recorrida por un objeto se calcula mediante la fórmula;

Distancia = velocidad * tiempo

Sustituyendo los valores en la fórmula, tenemos;

Distancia = 85 * 2

Distancia = 170 kilómetros

A continuación, convertiríamos el valor en kilómetros a metros;

Conversión:

1 kilómetros = 1000 metros

170 kilómetros = D metros

Multiplicación cruzada, tenemos;

D = 170 * 1000

D = 170,000 metros

Por lo tanto, habría viajado 170,000 metros después de 2 horas.

We know

[tex]\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}[/tex]

[tex]\\ \sf\longmapsto V_{AB}=2-2[/tex]

[tex]\\ \sf\longmapsto V_{AB}=0m/s[/tex]

The vector difference between the velocities of two bodies : the velocity of a body with respect to another regarded as being at rest  compare relative motion

[tex]Relative velocity $\left(\mathrm{V}_{\mathrm{AB}}\right)=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$$$\begin{aligned}&\longmapsto \mathrm{V}_{\mathrm{AB}}=2-2 \\&\longmapsto \mathrm{V}_{\mathrm{AB}}=0 \mathrm{~m} / \mathrm{s}\end{aligned}$$[/tex]

The relative velocity of an object with respect to another is the velocity with which one object moves with respect to another object. The unit of velocity can be referred to as the ratio of unit of distance and that of time. The SI unit of Relative velocity is meter per second.

The concept of absolute velocity is mainly used in turbomachinery design and defines the velocity of a fluid particle in relation to the surrounding, stationary environment. Together with the relative velocity (w) and the circumferential speed (u), it forms the velocity triangle.

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Answer:

About Transcript. Nuclear binding energy is the energy required to split an atom's nucleus into protons and neutrons. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference.

Explanation:

PLZ MARK AS THE BRAINLIEST

I NEED IT URGENTLY

HAVE A GOOD DAY

MAY GOD BLESS U

: )

Explanation:

a) copper

b) olive oil

Hope it helps✌✌

Answer:

Explanation: as,

v=f∧

v=200×10³×1.5×10³

v=2.7×10^7ms⁻¹

The velocity of the radio waves is 3 × 10⁸ ms⁻¹.

A radio station frequency = 200 KH

wavelength = 1.5 km

         Radio waves in the electromagnetic spectrum has the longest wavelength and it will always be below 300GHz. The radio waves can be generated with acceleration through some charged particles. Only through the transmitter via antenna the radio waves can gets transmitted. Radio waves can be used in all the electronic devices as mobile phones, radio communication, radars and navigations.

                               V = f λ

            Velocity,  v = ( 200 × 10³ ) × ( 1.5 × 10³ )

                               = ( 300 × 10⁶ )

                             v = 3 × 10⁸ ms⁻¹.

Hence, the velocity of the radio waves is 3 × 10⁸ ms⁻¹.

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Explanation:

tilting it will raise the height of its center of gravity.

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

therefore, work does 9900 N

Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).

Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.

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Answer:

The magnitude of the unknown force is 44.8 N.

Explanation:

The force can be found with Newton's second law:                

[tex] F = ma [/tex]

Where:

m: is the mass of the toy car = 5.5 kg

a: is the acceleration

F: is the force =?

We can calculate the acceleration with the following kinematic equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex] v_{f} [/tex]: is the final speed = 7.3 m/s

[tex] v_{0} [/tex]: is the initial speed = 2.1 m/s

d: is the distance traveled = 3 m

Hence, the acceleration is:

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(7.3 m/s)^{2} - (2.1 m/s)^{2}}{2*3 m} = 8.15 m/s^{2} [/tex]

Finally, the magnitude of the force is:

[tex]F = ma = 5.5 kg*8.15 m/s^{2} = 44.8 N[/tex]                                  

Therefore, the magnitude of the unknown force is 44.8 N.

I hope it helps you!                                

a.

Given,

height (h) = 100m

mass (m) = 5kg

acceleration due to gravitation (g) = 9.8ms^-2

Potential energy

= mgh

= 5kg × 9.8ms^-1 × 100m

= 4900 kgm²s^-2

= 4900 J

b.

Since the ball is not moving yet, its kinetic energy is 0.

c.

The total of potential and kinetic energy at every point of the journey is same, i.e., 4900 J.

Answer:

1. The efficiency of the radiator is 90 %

2. The efficiency of the torch is 65 %

3. The efficiency of the car is 40 %

4. The efficiency of the energy saver is 70 %

5. The efficiency of the speaker is 50 %

Explanation:

Efficiency = (Useful energy out ÷ Total energy in) × 100 J

1. Useful energy = 900 J

The total energy in = 1000 J

The efficiency of the radiator = ((900 J)/(1,000 J)) × 100 % = 90 %

2. Useful energy = 65 J

The total energy in = 100 J

The efficiency of the torch = ((65 J)/(100 J)) × 100 % = 65 %

3. Useful energy = 4,000 J

The total energy in = 10,000 J

The efficiency of the car = ((4,000 J)/(10,000 J)) × 100 % = 40 %

4. Useful energy = 700 J

The total energy in = 1,000 J

The efficiency of the energy saver = ((700 J)/(1,000 J)) × 100 % = 70 %

5. Useful energy = 50 J

The total energy in = 100 J

The efficiency of the speaker = ((50 J)/(100 J)) × 100 % = 50 %

Explanation:

Given that,

Mass of an object, m = 250 g = 0.25 kg

Spring constant, k = 48 N/m

The amplitude of the oscillation, A = 5.42 cm = 0.0542 m

1. At equilibrium,

ma = kx

Where

a is the acceleration of the object

So,

[tex]a=\dfrac{kx}{m}\\\\a=\dfrac{48\times 0.0542}{0.25}\\\\a=10.4\ m/s^2[/tex]

2. The maximum speed of the object is :

[tex]v=A\omega\\\\v=A\sqrt{\dfrac{k}{m}}\\\\v=0.0542\times \sqrt{\dfrac{48}{0.25}}\\\\v=0.75\ m/s[/tex]

Hence, this is the required solution.

Answer:

The elements are grouped into the different substances by color. As you can see, Lithium, Beryllium, Sodium, Magnesium, Aluminum, Potassium, and Calcium are metals out of the first 20 elements.

Hydrogen, Helium, Carbon, Nitrogen, Oxygen, Fluorine, Neon, Phosphorus, Sulfur, Chlorine, and Argon, are non-metals within the first 20 elements.

Boron and Silicon count as Metalloids in the Periodic Table (properties of both metals and non-metals)

reference- socatric q and a

Explanation:

Answer:

(11-13) - Metals

(14) - Metalloid

(15-18) - Non- metals

(19-20) - Metals

Explanation:

The elements from 11 to 13 are metals. They are Sodium(Na), Magnesium(Mg) and Aluminum(Al) respectively.

The element 14 is a metalloid. It’s Silicon(Si).

The elements from 15 to 18 are non metals. The 18th element is a noble gas known as Argon(Ar).

The elements from 15 to 17 are Phosphorus(P), Sulphur(S) and Chlorine(Cl).

The elements from 19 to 20 are metals. They are Potassium(K) and Calcium(Ca).

Answer:

3.8 x 10^-5

Explanation:

all you doing is putting it in the standard form I hope this help

Answer:

please do write it in english

Answer:

Explanation:

mass of linebacker, m = 111 kg

initial velocity, u = 1.9 m/s

mass of quarterback, m' = 82 kg

initial velocity, u' = 3 m/s

(a) Let they are moving in the same direction, the velocity is v after collision.

Use conservation of momentum

m u + m' u' = (m + m') v

111 x 1.9 + 82 x 3 = (111 + 82) v

v = 2.4 m/s

Let they are moving in opposite direction, the velocity is v after collision.

Use conservation of momentum

m u - m' u' = (m + m') v

111 x 1.9 - 82 x 3 = (111 + 82) v

v = - 0.18 m/s

Explanation:

If the mass of the inclined plane is large enough, could N ever be equal to mgcosθ. Reasons?

So far I've come up with: mgcosθ−N=ma meaning if N were to equal mgcosθ, ma=0. Since the surfaces are frictionless and since N is acting on the inclined plane too(it'll have a horizontal component), this isn't possible. Am I right? Where am I going wrong?

Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up

Answer:

The answer is below

Explanation:

Una partícula efectúa un MAS cuya ecuación es: x=0.3cos (2t + π/6) m. Determinar: Amplitud, frecuencia angular, fase inicial, periodo, frecuencia de oscilación y posición en t=0.25 s

Solution:

La ecuación de la onda es:

x = A cos (ω t + Ф), donde:

A = amplitud, ω = frecuencia angular = 2 π / T = 2 π f, Ф = fase inicial, f = frecuencia, T = período

Por lo tanto, comparando la ecuación de la onda con x = 0.3cos (2t + π / 6), obtenemos:

a) A = 0.3

b) ω = 2 rad / s

c) Ф = π / 6 rad

d) ω = 2π / T

2 = 2π / T

T = 3.14 s

e) ω = 2πf

2 = 2πf

f = 0.32 Hz

f) en t = 0.25 s:

x (t) = 0.3cos (2 * 0.25 + π / 6) = 0.22 m

Answer:

The Planck constant, or Planck's constant, is a fundamental physical constant denoted h, and is of fundamental importance in quantum mechanics. A photon's energy is equal to its frequency multiplied by the Planck constant. Due to mass–energy equivalence, the Planck constant also relates mass to frequency.

Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.

E = hf

[tex] \sf \: h = 6.626 \times {10}^{ - 34} \: kg \: {m}^{2} {s}^{ - 1} [/tex]

Momentum is conserved, so the sum of the momenta of the bullet and block before collision is equal to the momentum of the combined bullet-block system,

[tex]m_bv_b+m_wv_w = (m_b+m_w)v[/tex]

where v is the speed of the bullet-block system. The block starts at rest so it has no initial momentum, and solving for v gives

[tex]v = \dfrac{m_b}{m_b+m_w} v_b[/tex]

The total work W performed by the spring on the bullet-block system as it is compressed a distance x is

[tex]W = -\dfrac12kx^2[/tex]

where k is the spring constant, and the work done is negative because the restoring force of the spring opposes the bullet-block as it compresses the spring.

By the work-energy theorem, the total work done is equal to the change in the bullet-block's kinetic energy ∆K, so we have

[tex]W_{\rm total} = W = \Delta K[/tex]

The bullet-block starts moving with velocity v found earlier and comes to a stop as the spring slows it down, so we have

[tex]-\dfrac12kx^2 = -\dfrac12(m_b+m_w)v^2 \implies kx^2 = \dfrac{{m_b}^2}{m_b+m_w}{v_b}^2[/tex]

Solve for [tex]m_w[/tex]:

[tex]m_w=\dfrac1k\left(\dfrac{m_bv_b}x\right)^2-m_b[/tex]

[tex]m_w=\dfrac1{205\frac{\rm N}{\rm m}}\left(\dfrac{(0.0135\,\mathrm{kg})\left(245\frac{\rm m}{\rm s}\right)}{0.350\,\rm m}\right)^2-0.0135\,\mathrm{kg}\approx \boxed{0.422\,\mathrm{kg}}[/tex]