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otez555 [7]
3 years ago
5

Students are given some resistors with various resistances a battery with internal resistance

Physics
2 answers:
Hunter-Best [27]3 years ago
5 0
One of the primary reasons why a battery should have an internal resistance is that because it is connected to an electromotive source. This electromotive force that is caused by a potential difference would be used for driving the current to power the electrical loads the battery sustains.
balu736 [363]3 years ago
5 0

Answer:

Explanation

Explanation:

V  = emf − Ir ,

where r is the internal resistance and I is the current flowing at the time of the measurement.

I is positive if current flows away from the positive terminal. The larger the current, the smaller the terminal voltage.

Resistances is series add  the total resistances

while in parallel reduces .

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Volgvan

The bulbs will produce lesser light than their capacity, In short they will be dimmer because the the energy will get divided in the number of bulbs.

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Brainly malfunction?|<br><br> I'm not sure why but...
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Explanation:

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3 years ago
Which of the following is true regarding the properties of light?
Ivanshal [37]

Light having a dual nature and acting like both a wave and a particle is the correct statement in this scenario.

<h3>What is Light?</h3>

This refers to the electromagnetic radiation found in the electromagnetic spectrum that is perceived by the human eye and has a dual nature. It doesn't require a medium for its propagation unlike sound.

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6 0
1 year ago
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
a_sh-v [17]

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

         h’ = 1/9 h

6 0
3 years ago
(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
murzikaleks [220]

Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
3 years ago
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