Ask question

Ask question

All categories

2 years ago

7

Dos bolas, de masas mA = 40 g y mB =60 g, esta?n suspendidas como se observa en la figura. La bola ma?s ligera se jala en un a?n

gulo de 60° con respecto a la vertical y se libera. A) ?Cua?l es la velocidad de la bola ma?s ligera antes del impacto? B) ?Cua?l es la velocidad de cada bola despue?s de la colisio?n ela?stica? C) ?Cua?l sera? la altura ma?xima de cada bola despue?s de la colisio?n ela?stica?

1 answer:

Ymorist [56]2 years ago

3 0

MA=40 or something I really don’t know

You might be interested in

Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro

algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

=56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its

m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0

3 years ago

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

$J = - D \frac{\Delta c}{\Delta x}$

Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

$D= D_0 e^{\frac{E_d}{RT}}$

Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

Rearrange the equation to get the value of temperature,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

$\Delta x = 9.7*10^{-3}m$

$E_d = 82000J$

$D_0 = 6.5*10^{-7}m^2/s$

$J = 3.2*10^{-9}m^2/s$

$R= 8.31Jmol^{-1}K$

Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

$T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}$

$T=1118.07K=844\°C$

4 0

3 years ago

because lighting heats up the surrounding air slowly, it realationship to thunder is currently unknown true or false

NeTakaya

Answer:

False

Explanation:

8 0

2 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed

DanielleElmas [232]

Answer:

$W = 10.28\ mm$

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

$a sin \theta_1 = m\lambda$

$\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})$

m = 1

$\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})$

$\theta_1 = 0.1133^\circ$

Width of the first minima

$y_1 = L tan \theta_1$

$y_1 = 2.60\times tan( 0.11331)$

$y_1 = 5.14 \ mm$

Now, width of the central region

$W = 2 y_1$

$W = 2\times 5.14$

$W = 10.28\ mm$

8 0

3 years ago