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Nataly [62]
3 years ago
9

Electric Field Hockey

Physics
2 answers:
Tomtit [17]3 years ago
8 0

Answer:

A. The electric field lines point outward and the black puck moves backwards. From this I can see that I can't put the red charge in front of the black puck because it will go away from the goal.

B. The electric field lines point inward and the black puck moves forwards and backwards. The conclusion I made from this was the blue charge by itself will not be able to get into the goal.

C.The puck moves faster when the mass increases and moves slower when the mass decreases. The more mass, the more force that is applied to the puck.

D.  The relationship between the speed of the black puck and its distance from the blue charge is when the puck is further away from the blue charge, the slower the puck moves away from it.

Explanation:

mars1129 [50]3 years ago
5 0

Answer:

The positive charges point away from each other

Explanation:

Why?

Arrows point away from the positive charge and toward the

negative charge.

Explanation:

Suppose that you rubbed a balloon with a sample of animal fur such as a wool sweater or even your own hair. The balloon would likely become charged and its charge would exert a strange influence upon other objects in its vicinity. If some small bits of paper were placed upon a table and the balloon were brought near and held above the paper bits, then the presence of the charged balloon might create a sufficient attraction for the paper bits to raise them off the table. This influence - known as an electric force - occurs even when the charged balloon is held some distance away from the paper bits. The electric force is a non-contact force. Any charged object can exert this force upon other objects - both charged and uncharged objects. One goal of this unit of The Physics Classroom is to understand the nature of the electric force. In this part of Lesson 1, two simple and fundamental statements will be made and explained about the nature of the electric force.

Perhaps you have heard it said so many times that it sounds like a cliché.

Opposites attract. And likes repel.

These two fundamental principles of charge interactions will be used throughout the unit to explain the vast array of static electricity phenomena. As mentioned in the previous section of Lesson 1, there are two types of electrically charged objects - those that contain more protons than electrons and are said to be positively charged and those that contain less protons than electrons and are said to be negatively charged. These two types of electrical charges - positive and negative - are said to be opposite types of charge. And consistent with our fundamental principle of charge interaction, a positively charged object will attract a negatively charged object. Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other. That is, a positively charged object will exert a repulsive force upon a second positively charged object. This repulsive force will push the two objects apart. Similarly, a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other.

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6 0
3 years ago
A boat is subject to hydrodynamic drag forces of F = 40v 2 [N] where v is the magnitude of velocity in [m/s]. The boat’s mass is
stepladder [879]

Answer:

It has moved a distance, S = 25.9 m

Explanation:

F = 40 v²........(1)

F = mv\frac{dv}{dS}.........(2)

Equating (1) and (2)

-mv\frac{dv}{dS} = 40 v^2\\\\v\frac{dv}{dS} = \frac{-40 v^2}{m}  \\\\m = 450 kg\\v\frac{dv}{dS} = \frac{-40 v^2}{450}\\\\ \frac{dv}{v} = \frac{40}{450} dS\\

Integrate both sides:

v_1 = 1, v_2 = 10

\int\limits^{10}_1 {\frac{1}{v} } \, dv = \frac{-40}{450}  \int\limits^S_0  \, dS \\\\ln\frac{1}{10} = \frac{-40}{450} (S-0)\\\\S = \frac{-40}{450} ln(0.1)\\\\S = 25.9 m.

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3 years ago
Which of the following is a measurement of acceleration?
Tems11 [23]
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4 years ago
A train traveling at 27.5 m/s accelerates to 42.4 m/s over 75.0 s. What is the displacement of the train in this time period
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Answer:

2621.25 meters

Explanation:

First, write down what we are given.

Initial velocity = 27.5 m/s

Final velocity = 42.4 m/s

Time = 75 seconds

We need to look at the kinematic equations and determine which one will be best.  In this case, we need an equation with distance.  I am going to use v_{f}^{2} = v_{i}^{2} +2ad, but you can also use the other equation, x = v_{o}t+\frac{1}{2}at^{2}

We need to find acceleration.  To find it, we need to use the formula for acceleration: a = \frac{v_{f}-v_{i}}{t}.  Plugging in values, a = \frac{42.4-27.5}{75} = .199\ m/s^{2}

Next, plug in what we know into the kinematics equation and solve for distance.  42.4^{2} = 27.5^{2} + 2(.199)(d)\\d = 2621.25\ meters

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