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kakasveta [241]
3 years ago
12

I need help ASAP

Physics
1 answer:
ddd [48]3 years ago
3 0
I’m pretty sure the correct answer is plasma I am so so so so sorry if it’s incorrect
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If you place small pieces of tissue paper near the balloon in Picture 2, they would probably stick to the
pishuonlain [190]

Answer:

The balloon has more negative charges than positive charges. The tissue paper has a neutral charge.The static electricity pulls on the tissue paper lifting it up.

4 0
3 years ago
2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan
ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

5 0
3 years ago
A SMART car can accelerate from rest to a speed of 28 m/s in 20s. What distance does it travel in this time?
sashaice [31]
Speed= distance/ time so distance = speed * time= 28 * 20= 560m .. so the answer is 560m.. l hope it helped :)
6 0
2 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
PLS HELP WILL MARK BRAINLIEST
MariettaO [177]

Answer:

2.835 Watts

Explanation:

P = I²R

P = 1.5² × 1.26

P = 2.835 Watts

3 0
3 years ago
Read 2 more answers
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