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3 years ago

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Two light bulbs are being used in a lab experiment with a single battery. Bulb A has half the resistance as Bulb B. Which bulb i

s the dimmest in series? Which bulb is the brightest in parallel? Explain.

1 answer:

Naya [18.7K]3 years ago

3 0

Answer and Explanation:

It is given that bulb A has the half the resistance as B

WHEN THEY ARE CONNECTED IN SERIES : In series circuit we know that current is same in the whole circuit

The power is given by $P=i^2R$ the expression it is clear that the bulb which has more resistance consume more power as current is same, and the bulb which consume more power will be brightest

It is given that resistance of bulb A is half the resistance of B So bulb A will be dimmest in this case

WHEN THEY ARE CONNECTED IN PARALLEL : In parallel circuit voltage is same across the circuit

The power is given by $P=\frac{V^2}{R}$ from the expression it is clear that the bulb which has least resistance will consume more power and so will be brightest

As the resistance of bulb A is less so it will be brightest in this case

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How does a parallel circuit differ from a series circuit?

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A series circuit has only one path for current,
while a parallel circuit has more than one.

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3 years ago

a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th

Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

And solving for $F_f$ , we find the force of friction:

$F_f = F-ma=500-(50)(4)=300 N$

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2 years ago

A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

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3 years ago

An isotope has the same ________, but a different ________ .

Lunna [17]

An isotope is one of the forms of an element that has the same number of protons, but a different number of neutrons.

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3 years ago

Read 2 more answers

Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a

Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

Vf = √( 2 g X)

Vf = √(2 9.8 1)

Vf = 4.43 m/s

This is the speed with which it reaches the ground

Having the final speed we can find the time

Vf = Vo + g t

t = Vf / g

t = 4.43 / 9.8

t = 0.45 s

This is the time of fall of the body to touch the ground

3 0

3 years ago