How is the resistor of a 40v lamp higher then the resistor of a 60v lamp?

Answers

Answer 1

Answer: Power dissipation = (voltage across the component)² / (resistance of the component)

Since the resistance is in the denominator of the fraction in this formula for the
quantity of power dissipated, you can see that when the supply voltage is constant,
the smaller resistance dissipates more power.

So the 40w lamp has higher resistance than the 60w bulb.

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Two objects with different masses collide and bounce back after an elastic collision. Before the collision, thetwo objects were moving at velocities equal in magnitude but opposite in direction. After the collision,
A. the less massive object had gained momentum.
B. the more massive object had gained momentum.
C. both objects had the same momentum.
D. both objects lost momentum.

Answers

The more massive object will lose momentum after collision while the less massive object will gain momentum after collision.

Let the mass of the first object = m₁
Let the mass of the second object = m₂
let the initial velocities of the two objects = u
let the final velocity of the first object after collision = v₁
Let the final velocity of the second object after collision = v₂

Apply theprinciple of conservation of linear momentum for elastic collision;

let the heavier object = m₁

m₁u + m₂(-u) = m₁(-v₁) + m₂v₂

m₁u - m₂u = -m₁v₁ + m₂v₂

where;

m₁u and m₂u are initial momentum of both objects before collision

m₁v₁ and m₂v₂ are final momentum of both objects after collision

Thus, from the equation above we can conclude the following, the more massive object will lose momentum after collision while the less massive object will gain momentum after collision.

Learn more here:brainly.com/question/11214803

Answer:A

Explanation:

Given

mass of two objects are $m_1$ and $m_2$

taking $m_1>m_2$

Suppose $u_1$ and $u_2$ are the velocities of $m_1$ and

$u_1=u$ and $u_2=-u$

therefore after elastic collision velocity of $m_1$ and $m_2/[tex] are [tex]v_1$ and $v_2$

$v_1=(m_1-m_2)/(m_1+m_2)\cdot u_1+(2m_2)/(m_1+m_2)\cdot u_2$

$v_2=(2m_1)/(m_1+m_2)\cdot u_1-(m_1-m_2)/(m_1+m_2)\cdot u_2$

for $u_1$ and $u_2$

$v_1=(m-m)/(m+m)\cdot u+(2m)/(m+m)\cdot (-u)$

$v_1=(m_1+m_2)/(m_1+m_2)\cdot u=u$

$v_2=(2m)/(m+m)\cdot u-(m-m)/(m+m)\cdot (-u)$

$v_2=(3m_1-m_2)/(m_1+m_2)\cdot u$

$v_2=u+2\cdot (m_1-m_2)/(m_1+m_2)\cdot u$

so velocity of mass $m_2$ is more as compared to $m_1$

so less massive object gained some momentum

A thin spherical glass shell in air is filled with an unknown liquid. A horizontal parallel light beam is incident on the shell and it is observed that the light is brought to focus on the surface of the shell directly opposite of the incoming beam. What is the refractive index of the liquid?ie. the answer is 2- I just don't know how they got there

Answers

I have the exact same question, any chance you figured it out since you posted this?

The four terrestrial planets that have solid, rocky surfaces are A. Mars, Venus, Earth, and Jupiter.
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C. Mercury, Venus, Jupiter, and Mars.
D. Mercury, Venus, Earth, and Mars

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The four terrestrial planets that hace solid, rocky surfaces are (D.) Mercury, Venus, Earth, and Mars.

Check out pages 804-808 in the Glencoe Earth Science textbook provided by PennFoster. That way you can learn about the answer. :)

Which of the following accurately describes properties of valence?A. The smaller the number of electrons an atom has to borrow or to lend, the greater the activity of the atom.
B. Nonmetallic elements tend to have a positive valence and tend to be electron borrowers.
C. Metals tend to have a negative valence and tend to be electron borrowers.
D. The greater the number of electrons an atom has to borrow or to lend, the greater the activity of the atom

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Answer:

Explanation:

The strength of gravitational force is dependent on the masses of and distance between two objects. a. True
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Answers

Truer words are seldom written.

Help me on this question? log x-log 9=1

Answers

$\log { x } -\log { 9 } =1\n \n \log { \left( \frac { x }{ 9 } \right) } =1\n \n \log _( 10 ){ \left( \frac { x }{ 9 } \right) } =1\n \n { 10 }^( 1 )=\frac { x }{ 9 } \n \n 9\cdot 10=x\n \n x=90$

This is because:

$\log _( a ){ \left( \frac { x }{ p } \right) } \n \n =\log _( a ){ \left( \frac { { a }^( m ) }{ { a }^( n ) } \right) } \n \n =\log _( a ){ \left( { a }^( \left( m-n \right) ) \right) } \n \n =\left( m-n \right) \cdot \log _( a ){ a } \n \n =m-n\n \n =\log _( a ){ x } -\log _( a ){ p } \n$

   log(x) - log(9) = 1

Subtracting the logs of numbers gives you the log of
the quotient of the numbers.

   Log(x) - log(9) is the log of (x/9).

So the equation says: log (x/9) = 1

Raise 10 to the power of each side: 10^(log of x/9) = 10^1

But 10^(log of x/9) is x/9, and 10^1 is 10.

So x/9 = 10

   x = 90

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