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Ilia_Sergeevich [38]
4 years ago
10

Lillian (mass 40.0 kg) standing at rest on slippery ice catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Wha

t is the speed of Lillian and her dog after the catch?
Physics
1 answer:
mestny [16]4 years ago
3 0

Answer:

0.82 m/s

Explanation:

From the law of conservation of momentum,

Total momentum = Final momentum.

mu+m'u' = V(m+m')......................... equation 1

Where m = mass of Lillian, u = initial velocity of Lillian before she catches the dog, m' = mass of the dog, u' = initial velocity of the dog, V = velocity of Lillian and the dog.

make V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Given: m = 40 kg, m' = 15 kg, u = 0 m/s (Lillian was standing at rest), u' = 3.0 m/s.

Substitute into equation 2

V = (40×0+15×3)/(40+15)

V = (0+45)/55

V = 45/55

V = 0.82 m/s

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Identify the names, charges, and location of three types of subatomic particles that make up an atom.
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Answer:Proton (charge of +e, in the nucleus), Neutron (0 charge, in the nucleus), and Electron (charge of –e, outside the nucleus).Nov 13, 2015

Explanation: YA

3 0
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A certain string can withstand a maximum tension of 39. N without breaking. A child ties a 0.43 kg stone to one end and, holding
frez [133]

Answer:

Bottom of the circle.

Explanation:

At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.

6 0
3 years ago
If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Klio2033 [76]

Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>
  • m= 18.02597
  • c=3.00×10⁸ms¯¹

<h3><u>required </u><u>value</u></h3>
  • Eb=?

<h3><u>Solution:</u></h3>
  • Eb=(Δm)c²
  • Eb=(18.02597)*(3.00*10⁸ms¯¹
  • Eb=5.407791*10⁹J

8 0
2 years ago
The de Broglie wave is produced only by sub atomic particle and photon. O True O False
Alina [70]

Answer:

True

Explanation:

Matter can be in the form of a particle or a wave. This is known as the dual nature of matter. This concept was proposed by Louis de Broglie and was named after him. This phenomenon has been observed for all the elementary particles.

The de Broglie wavelength is given by

\lambda=\frac{h}{p}=\frac{h}{mv}

Where

h = Planck's constant

p = Particles momentum

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8 0
3 years ago
The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared
velikii [3]

Answer:

\displaystyle a(5)=-32

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

\displaystyle v(t)=\frac{df}{dt}

And the acceleration is

\displaystyle a(t)=\frac{dv}{dt}

Or equivalently

\displaystyle a(t)=\frac{d^2f}{d^2t}

The given height of a projectile is

f(t)=-16t^2 +238t+3

Let's compute the speed

\displaystyle v(t)=-32t+238

And the acceleration

\displaystyle a(t)=-32

It's a constant value regardless of the time t, thus

\boxed{\displaystyle a(5)=-32}

3 0
3 years ago
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