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Sauron [17]
3 years ago
14

I) A 100 W and 60 W bulb are joined in series and connected to the mains. Which bulb

Physics
1 answer:
kolbaska11 [484]3 years ago
3 0

Answer:

i) 60 W

ii) 100 W

Explanation:

In each case, the bulb that dissipates the most power is the bulb that glows brighter.  Power is voltage times current (P = VI).  Using Ohm's law, we can rewrite this as P = I²R or P = V²/R.

Bulbs are rated at a certain power for a certain voltage.  P = V²/R, so the bulb with the lower resistance will have the higher power rating.  Therefore, the 100 W bulb has a lower resistance than the 60 W bulb.

i) They are in series, so they have the same current.  P = I²R, so the bulb with the higher resistance will glow brighter.  That's the 60 W bulb.

ii) They are in parallel, so they have the same voltage.  P = V²/R, so the bulb with the lower resistance will glow brighter.  That's the 100 W bulb.

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A jet on an aircraft carrier can be launched from rest to 40 m/s in 2 seconds. What is the acceleration of the aircraft? Show st
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Answer:

a=20\ m/s^2

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 40 m/s

Time, t = 2 s

We need to find the acceleration of the aircraft. We know that, acceleration is equal to the rate of change of velocity. So,

a=\dfrac{v-u}{t}\\\\a=\dfrac{40-0}{2}\\\\=20\ m/s^2

So, the acceleration of the aircraft is 20\ m/s^2.

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3 years ago
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Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

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The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

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