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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning

zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

$L_{i}=L_{f}$

Here, = the system's angular momentum before the collision

$L_{i}$ = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here, = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

4 0

1 year ago

Read 2 more answers

What from the following list of statements about vectors is definitely true? (section 3.3) The magnitude of a vector can be smal

Mashcka [7]

Answer:

"Magnitude of a vector can be zero only if all components of a vector are zero."

Explanation:

"The magnitude of a vector can be smaller than length of one of its components."

Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.

"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."

False. Magnitude of a vector is always positive.

"Magnitude of a vector can be zero if only one of components is zero."

Wrong. For the magnitude of a vector to be zero, all components must be zero.

"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."

Wrong. The magnitude of a vector depends on all components, not only the X component.

"Magnitude of a vector can be zero only if all components of a vector are zero."

True.

6 0

3 years ago

Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d

uysha [10]

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

$Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}$

3 0

3 years ago

What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to e⃗

AleksandrR [38]

Electric potential energy of a dipole is given as

$U = - P.E$

$U = - PE cos\theta$

so change in the potential energy is given by

$U_f - U_i = (-PEcos\theta_2)- (-PEcos\theta_1)$

$\Delta U = PE (cos\theta_1 - cos\theta_2)$

here initially it was parallel to electric field while finally it is perpendicular to the electric field

so we have

$\Delta U = PE(cos 0 - cos 90)$

$\Delta U = PE$

so above is the change in potential energy of dipole

8 0

3 years ago