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2 years ago

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Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c

ollision. What is the total momentum of the system containing the two balls before the collision? What about after the collision?

1 answer:

Rus_ich [418]2 years ago

5 0

Answer:

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

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1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

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2 years ago

Which one of the following statements concerning kinetic energy is true? a The kinetic energy of an object always has a positive

hodyreva [135]

a The kinetic energy of an object always has a positive value.

Explanation:

The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Let's now analyze each statement:

a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term $v^2$ is always positive as well, so the kinetic energy is always positive.

b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, $K\propto v^2$ .

c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.

d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.

e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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2 years ago

A microwave oven operates at 2.60 ghz . what is the wavelength of the radiation produced by this appliance?

cestrela7 [59]

<span>10.3 cm The wavelength will be the distance that light travels in 1 second divided by the frequency of the radiation. Since the over operates at 2.60 ghz, the frequence is 2.6 billion times per second, or 2.60 x 10^9. The speed of light is defined as 299792458 m/s exactly. So 299792458 m/s / 2.60 x 10^9 1/s = 0.10337671 m = 10.337671 cm Since we only have 3 significant digits, the answer rounds to 10.3 cm</span>

8 0

2 years ago

A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh

Jet001 [13]

Answer:

heat used to rise temperature pan = 30.1%

heat used to rise temperature water = 69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature = 80.0ºC

to find out

heat used to rise temperature of pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0

2 years ago