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vladimir2022 [97]
4 years ago
8

Which statement is TRUE? Question 13 options: In a series circuit, as light bulbs are added, the bulbs dim. In a parallel circui

t, if one bulb goes out, all of the bulbs will go out. In a series circuit, if one bulb goes out the other bulbs will stay lit. In a series circuit, the resistors do not share the current.
Physics
2 answers:
koban [17]4 years ago
8 0

Answer:

In a series circuit, as light bulbs are added, the bulbs dim

Explanation:

In a series circuit, as light bulbs are added, the bulbs dim because of two reasons.

1. In a series circuit, there is only one path for the current to flow and as more bulbs are added, their resistance gets added too, which results in less current.

2. In a series circuit, voltage across each bulb is different, as more bulbs are added, the voltage across each bulb becomes less and less.

Since the brightness of the bulbs is dependent on voltage and current

P = V*I

Both of these quantities are decreasing as the number of bulbs are increasing therefore, resulting in less power dissipation and eventually, less brightness.

Other options:

  • In a parallel circuit, if one bulb goes out, all of the bulbs will go out. False, In a parallel circuit, if one bulb goes out, then there is still a path left for the current to flow, therefore other bulbs are not affected.
  • In a series circuit, if one bulb goes out the other bulbs will stay lit    False, In a series circuit, there is only one path for the current to flow, when one bulb goes out then it blocks the way of the current and therefore, other bulbs get affected and will not stay lit.
  • In a series circuit, the resistors do not share the current.                    False, In a series circuit, the resistors share the same current since there is only path for the current to flow therefore, each resistor get the same current.
VikaD [51]4 years ago
3 0
The only statement which is true is:
<span>In a series circuit, the resistors do not share the current.

In fact, when resistors are put in series, the current I crosses all the resistors with same intensity, and so the resistors are all crossed by the same value of the current.</span>
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The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
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8 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
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