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BlackZzzverrR [31]
2 years ago
7

How is momentum conserved is a Newton's cradle when one steel ball hits the other

Physics
1 answer:
Vikentia [17]2 years ago
4 0

Answer:

Newton's Cradle experiment perfectly demonstrates the law of conservation of momentum which states that in a closed system, momentum before the collision is equal to momentum after the collision of the system.

As the first ball swings in the air, it gains momentum. When it strikes the second ball, it loses momentum and second ball gains equal amount of momentum. The second ball transfers the momentum to third, then fourth and till the last. The last ball when gains the same momentum swings up in the air. This continues. This experiment is done in drag free condition. This means there is no loss of momentum or opposing forces present.


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Which of the following is an obstacle to creating computer-based models for tracking a hurricane?
iren [92.7K]

Answer:

4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck

5 0
3 years ago
When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc
gtnhenbr [62]

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07  = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

7 0
2 years ago
Formula for the distance (d) is given by d = rate*time. For example if you are traveling at 60 mph for 3 hours the distance trav
babunello [35]

Explanation:

Distance covered by the particle is given by:

Distance (d) = rate (v) × time (t)                

Speed of Mary, v₁ = 50 mph

Speed of Jim, v₂ = 60 mph

It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.

Distance travelled by Jim, d₁ = (60t + 10)

Distance travelled by Mary, d₂ = 50t

The distance between Mary and Jim is greater than or equal to 100 miles.

60t+10-50t\ge100

10t\ge90

t\ge9\ h

So, Jim takes is 9 hours more than Mary to cover same distance. Hence, this is the required solution.

7 0
2 years ago
Air resistance is an example of what type of friction?
Sav [38]
Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!
4 0
3 years ago
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
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