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Nezavi [6.7K]
3 years ago
9

What is the proper battery cable connection when jumping two automotive batteries? (a) negative to negative / positive to positi

ve (b) negative to positive / negative to positive (c) it doesn't matter on 12-volt automotive batteries?
Physics
1 answer:
Vladimir [108]3 years ago
3 0
<span>The proper </span><span>battery cable connection when jumping two automotive batteries is :  </span><span>(a) negative to negative / positive to positive. 

</span><span>Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. </span>

<span>Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.
</span>
<span>Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.</span>


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Which is an example of a primitive plant?
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Answer:

Non-flowering plants like mosses, horsetails, ferns, clubmosses, ginkgos, and cycads

Explanation:

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Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

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3 years ago
a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi
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5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

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Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

        Momentum = mass x velocity

 Since the speed of the ball is unknown:

           Velocity  =  \frac{momentum }{mass}

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                           = 5.2m/s

The speed of the baseball before it lands is 5.2m/s

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Answer: Two men who helped in the development of psychology in Germany was Wilhelm Wundt, and in United States was William James.

Explanation:

Wilhelm Wundt was a German scientist, who was the first person who introduced psychology. He studied the conscious experience as a part of scientific study. It emphasis the identification of the components of the consciousness. Introspection is the process through which conscious experiences can be analyzed.

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4 years ago
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