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Grace [21]
3 years ago
10

Integrated Concepts. A flashing lamp in a Christmas earring is based on an RC discharge of a capacitor through its resistance. T

he effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. What energy (J) does it dissipate
Physics
2 answers:
Anton [14]3 years ago
5 0

Power = (energy) / (time)

0.5 W = (energy) / (0.25 sec)

Multiply each side by (0.25 sec):

Energy = (0.5 W) x (0.25 sec)

Energy = 0.125 W-sec

<em>Energy = 0.125 Joule</em>

We can ignore the Christmas trinket, the RC discharge, the capacitor, the resistor, and the voltage.  We don't need any of them in order to calculate the energy of each flash.  They're only there to confuse us, distract us, and see whether we're paying attention.

SashulF [63]3 years ago
4 0

Answer:

Energy=0.125J

Explanation:

To find

The dissipated Energy

Given data

Duration of flash t=0.250s

Power P=0.500 W

Voltage V=3.00V

As we know that

Power=\frac{Energy}{time} \\Energy=Power*time\\Energy=0.500W*0.250s\\Energy=0.125J

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A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.
kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

   Therefore, average speed of the trip = \frac{\text{Total distance covered}}{\text{Total time taken}}

    From the graph attached,

   Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

                                           = 90 meters

   Total time taken = 60 seconds

    Average speed = \frac{90}{60}

                               = 1.5 meter per second

2). Velocity is a vector quantity.

    Therefore, average velocity = \frac{\triangle d}{\triangle t}

                                                   = \frac{d_{60}-d_0}{60-0}

                                                   = \frac{90-0}{60-0}

                                                   = 1.5 meter per second                        

7 0
3 years ago
A web page designer creates an animation in which a dot on a computer screen has a position of r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i
VMariaS [17]

Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

5 0
3 years ago
Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac
Tems11 [23]
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
f_n = n f_1
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz
while the second overtone (3rd harmonic) is
f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz

Similarly, for the second string with fundamental frequency f_1 = 523 Hz, the first overtone is
f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz
and the second overtone is
f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz

(b) The fundamental frequency of a string is given by
f=  \frac{1}{2L}  \sqrt{ \frac{T}{\mu} }
where L is the string length, T the tension, and \mu = m/L is the mass per unit of length. This part  of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them m_{196}, the mass of the string of frequency 196 Hz, and m_{523}, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
\frac{523 Hz}{196 Hz} =  \frac{ \frac{1}{2L}  \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L}  \sqrt{ \frac{T}{m_{196}/L} }}
and since L and T simplify in the equation, we can find the ratio between the two masses:
\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1

(c) Now the tension T and the mass per unit of length \mu is the same for the strings, while the lengths are different (let's call them L_{196} and L_{523}). Let's write again the ratio between the two fundamental frequencies
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }} 
And since T and \mu simplify, we get the ratio between the two lengths:
\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them T_{196} and T_{523}. Let's write again the ratio of the frequencies:
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}
Now m and L simplify, and we get the ratio between the two tensions:
\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14
7 0
3 years ago
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