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Salsk061 [2.6K]
3 years ago
7

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

r. (Throughout, assume that each battery has negligible internal resistance.) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that R remains constant when the power consumption changes.
Physics
1 answer:
ch4aika [34]3 years ago
5 0

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

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A 4.50-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Citrus2011 [14]

Answer:

a)  μ = 0.1957 , b) ΔK = 158.8 J , c)    K = 0.683 J

Explanation:

We must solve this problem in parts, one for the collision and the other with the conservation of energy

Let's find the speed of the wood block after the crash

Initial moment. Before the crash

            p₀ = m v₁₀ + M v₂₀

Final moment. Right after the crash

           pf = m v_{1f} + M v_{2f}

           

The system is made up of the block and the bullet, so the moment is preserved

           p₀ = pf

          m v₁₀ = m v_{1f} + M v_{2f}

          v_{2f} = m (v₁₀ - v_{1f}) / M

          v_{2f} = 4.5 10-3 (400 - 190) /0.65

          v_{2f} = 1.45 m / s

Now we can use the energy work theorem for the wood block

Starting point

                Em₀ = K = ½ m v2f2

Final point

                Emf = 0

                W = ΔEm

               - fr x = 0 - ½ m v₂₂2f2

The friction force is

               fr = μN

     

With Newton's second law

               N- W = 0

               N = Mg

We substitute

               -μ Mg x = - ½ M v2f2

                μ = ½ v2f2 / gx

Let's calculate

            μ = ½ 1.41 2 / 9.8 0.72

            μ = 0.1957

b) let's look for the initial and final kinetic energy

           K₀ = 1/2 m v₁²

           K₀ = ½ 4.50 10⁻³ 400²

           K₀ = 2.40 10²  J

           Kf = ½ 4.50 10⁻³ 190²

           Kf = 8.12 10¹  J

Energy reduction is

              K₀ - Kf = 2.40 10²- 8.12 10¹

              ΔK = 158.8 J

c) kinetic energy

              K = ½ M v²

              K = ½ 0.650 1.45²

              K = 0.683 J

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