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BARSIC [14]
3 years ago
7

How do resistors in parallel affect the total resistance?

Physics
1 answer:
4vir4ik [10]3 years ago
4 0

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

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A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.
astra-53 [7]

Answer:

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be  -0.5026m/sec^2

Explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

S=ut+\frac{1}{2}at^2

40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2

7.45u+27.751a=40-----eqn 1

According to first equation of motion

v = u + at

So 3.5=u+7.45a-----eqn2

Solving equation 1 and 2

a = -0.5026m/sec^2

And u = 7.244 m /sec

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be

-0.5026m/sec^2

6 0
4 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
3 years ago
What is the difference between kinetic and potential energy and how do they work?
Iteru [2.4K]

To explain, I will use the equations for kinetic and potential energy:

PE = mgh\\KE = \frac{1}{2}mv^{2}

<h3>Potential energy </h3>

Potential energy is the potential an object has to move due to gravity.  An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>.  If gravity = 0 or height = 0, there is no potential energy.  Example:

An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2).  What is the object's gravitational potential energy?  <u>(answer: 5*5*9.8 = 245 J</u>)

(gravitational potential energy is potential energy)

<h3>Kinetic energy</h3>

Kinetic energy is the energy of an object has while in motion.  An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary).  An example:

An object of 5 kg is moving at 5 m/s.  What is the object's kinetic energy?  (<u>answer: 5*5 = 25 J</u>)

<h3>Kinetic and Potential Energy</h3>

Sometimes, an object can have both kinetic and potential energy.  If an object is moving (kinetic energy) and is above the ground (potential), it will have both.  To find the total (mechanical) energy, you can add the kinetic and potential energies together.  An example:

An object of 5 kg is moving on a 5 meter table at 10 m/s.  What is the objects mechanical (total) energy?  (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)

7 0
3 years ago
In static electricity what are the subatomic particles, their charges, their location in the atom
enot [183]
Electrons - have a negative charge and orbit the nucleus
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6 0
4 years ago
If this did not happen, what would be the approximate force on an eardrum of area 0.22 cm2 if a change in altitude of 1500 m tak
jeyben [28]

Complete Question

When you ascend or descend a great deal when driving in a car yours ears "pop," which means that the pressure behind the eardrum is being equalized to that outside. If this did not happen, what would be the approximate force on an eardrum of area .50 cm2 if a change in altitude of 950 m takes place?

Answer:

The value is F    = 0.60 \  N

Explanation:

From the question we are told that

   The area of the ear drum is  A = 0.5 \  cm^2 = 0.50 *10^{-4} \  m^2

    The change in altitude is  \Delta d  = 950 \  m

Generally the change in pressure is mathematically represented as

       \Delta P = \frac{F}{A}

This can also be mathematically represented as

      \Delta P = \rho * g *  \Delta d

So

       \frac{F}{A}     = \rho * g *  \Delta d

=>    F    = \rho * g *  \Delta d  *  A

=>    F    = \rho * g *  \Delta d  *  A

Here \rho is the density of dry air with value  \rho =  1.29 \ kg /m^3

So

     F    = 1.29 * 9.8 *  950  *  0.50 *10^{-4}

=> F    = 0.60 \  N

3 0
3 years ago
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