Can someone answer and explain these questions?
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Answer:
a. The ratio of their resistance is 2783:64
b. The ratio of their energy is 4:23
c. The charge on the first bulb is 5.75 C
The charge on the second bulb is C
Explanation:
The voltage on one of the electric bulbs, V₁ = 40 volts
The power rating of the bulb, P₁ = 230 w
The voltage on the other electric bulbs, V₂ = 110 volts
The power rating of the bulb, P₂ = 40 w
a. The power is given by the formula, P = I·V = V²/R
Therefore, R = V²/P
For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96
The resistance of the second bulb, R₂ = 110²/40
The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64
∴ The ratio of their resistance, R₂:R₁ = 2783:64
b. The energy of a bulb, E = t × P
Where;
t = The time in which the bulb is powered on
∴ The energy of the first bulb, E₁ = 230 w × t
The energy of the second bulb, E₂ = 40 w × t
The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23
∴ The ratio of their energy, E₂:E₁ = 4:23
c. The charge on a bulb, 'Q', is given by the formula, Q = I × t
Where;
I = The current flowing through the bulb
From P = I·V, we get;
I = P/V
For the first bulb, the current, I = 230 w/40 V = 5.75 amperes
The charge on the first bulb per second (t = 1) is therefore;
Q₁ = 5.75 A × 1 s = 5.75 C
The charge on the first bulb, Q₁ = 5.75 C
Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = C
The charge on the second bulb, Q₂ = C.
d. The question has left out parts
The apparent weight of a 1.1 g drop of water is 4.24084 N.
<h3>What is Apparent Weight?</h3>- According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
- By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.
Solution:
N = Speed of rotation = 1250 rpm
D = Diameter = 45 cm
r = Radius = 22.5 cm
M = Mass of drop = 1.1 g
Angular speed of the water =
Apparent weight is given by
= 4.24084 N
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Question:
The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?