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scoray [572]
3 years ago
13

The light bulb transfers electrical energy into light. What is one type of energy that is also generated that is NOT a desired e

ffect?
A)
chemical
B)
heat
C)
mechanical
D)
sound
Physics
2 answers:
klasskru [66]3 years ago
7 0

Answer:

B

Explanation:

heat

Troyanec [42]3 years ago
5 0

Light bulbs get hot !

The heat is not a desired effect, and the light bulb is not very efficient,
if you selected a light bulb to serve as a source of light.  But if chose a
light bulb to use in a place where you need heat, then the heat you get
from it is very desirable, and the light bulb is very efficient.  It all depends
on your point of view.
 
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Answer:

answer is A

Explanation:

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3 years ago
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find
storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

$f_{m_2}= 95 \ Hz$

Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

When n = 2,

$f_{o_1}= \pm 65 \pm 2(245),\ \  f_{o_2}=\pm 95 \pm 2(245)$

$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt
sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

f_{1}=\dfrac{v}{4l}

Put the value into the formula

f_{1}=\dfrac{344}{4\times0.11}

f_{1}=781.81\ Hz

f_{1}=0.782\ kHz

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

f_{1}=\dfrac{v}{4(\dfrac{L}{2})}

Put the value into the formula

f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}

f_{1}=1563.63\ Hz

f_{1}=1.563\ kHz

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0
4 years ago
Which of the following is a push or pull?
Nutka1998 [239]

Answer: A. Newton

Explanation:

A force is a push or pull exerted on an object and a newton is the type of measurement for a force.

4 0
4 years ago
Read 2 more answers
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