Many flashlights are powered by a stack of batteries that are slid into the end of the flashlight one after another. Suppose you
1 answer:
Answer:
The current through the bulb when the flashlight is turned on is 0.55 A.
Explanation:
Given that,
You have a flashlight that takes four ideal 1.5 V batteries, voltage of the flashlight, V = 6 V
The resistance of bulb, R = 11 ohms
We need to find the current through the bulb when the flashlight is turned on. As per Ohms's law :
V = IR
So, the current through the bulb when the flashlight is turned on is 0.55 A. Hence, this is the required solution.
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Answer:
a) 1.301 kg/s
b) 0.001301 m³/s
c) V₁ = 6.505 m/s, V₂ = 1.626 m/s
d) 118.93 kPa
Explanation:
Given:
The number of cans = 220
The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³
time = 1 minute = 60 seconds
gauge pressure at point 2, P₂ = 152 kPa
b) Thus, the volume flow rate, Q = Volume/ time
Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s
a) mass flow rate = Volume flow rate × density
since it is mostly water, thus density of the drink = 1000 kg/m³
thus,
mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s
c) Given:
Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²
Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²
also,
Q = Area × Velocity
thus, for point 1
0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)
or
V₁ = 6.505 m/s
for point 2
0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)
or
V₂ = 1.626 m/s
d) Applying the Bernoulli's theorem between the points 1 and 2 we have
or
on substituting the values in the above equation, we get
it is given that point 1 is above point 2 thus, y₂ -y₁ is negative
or
thus, gauge pressure at point 1 is 118.93 kPa