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lisabon 2012 [21]
3 years ago
15

Can someone please help me on this!!!

Physics
1 answer:
elena-s [515]3 years ago
6 0

Answer:

IDK

Explanation:

IDK

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How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can
lord [1]

Answer:

I'm not sure..but please refer to your teacher later.

Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.

Explanation;

  • smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
  • based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
  • now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
  • high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
  • Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly
6 0
3 years ago
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Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T
Westkost [7]

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

KE=\dfrac{1}{2}mv^2

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

3 0
3 years ago
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Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste
Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

W=0

We need to calculate the system's potential energy

Using thermodynamics first equation

\Delta U=W-\Delta E

Put the value into the formula

\Delta U=0-147

\Delta U=-147\ J

Hence, The system's potential energy is -147 J.

5 0
3 years ago
When electrical energy is being used by an electric light, what really happens to the energy?
vitfil [10]

Energy is not created and not  destroyed it will only change form


So heres your answer ; It is given off as other forms of energy/light and heat !!


=answer 2nd one

7 0
3 years ago
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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
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