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xxMikexx [17]
3 years ago
15

An alpha particle is identical to a:

Physics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

B-hydrogen nucleus

Explanation:

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A scientist heard a flexible container full of neon gas. What will be likely to happen to the container as the gas absorbed heat
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A bird, accelerating from rest at a constant rate, experiences a displacement of 28 m in 11 s. what is the final velocity after
frutty [35]
Let's see what variables we've got first. Hmmm. We have:

Displacement, d = 28 m
Time taken, t = 11 s
Initial velocity, u = 0 m/s (at rest)

And now we need to find the final velocity, v. Among the 4 (or 5) equations of motions, there's no equation that will let us simply plug in the values and give an answer sigh. But fear not! We'll do it in steps.

I'm going to pick one of the motion equation to find more information:

d = ut +  \frac{1}{2} a {t}^{2}

I know everything except for a in this one, so I I'll use this! After plugging in values, I get a = 0.4628 m/s^2.

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v = u + at

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4 0
3 years ago
12. What is the photoelectric effect, and what experimental evidence led scientists
anygoal [31]

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Explanation:

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6 0
3 years ago
Find time when boy catches the girl or when they are at their closest separation..
soldier1979 [14.2K]

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\

\green{\underline{\bold{Given :}}}  \\ \\   \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\  \\   \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\  \\  \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy =  {1 \: m/s}^{2}} \\  \\  \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl=  {2\: m/s}^{2} }\\  \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\  \\ \red{\underline{\bold{To \: Find :}}} \\ \\  \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl  }\\

<u>According to given question</u> :

\\ \green{\star} \:  \text{Using \: relative \: motion \: method} \\  \\  \green{ \circ} \:  \tt Net \: velocity = 50 - 30 = 20 \: m/s \\  \\ \green{ \circ} \:  \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2}  \\\\  \\  \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\  \tt\purple{:  \implies s = ut +  \frac{1}{2}  {at}^{2}}  \\  \\ \tt\green{:  \implies 150 = 20 \times t +  \frac{1}{2}  \times  -1 \times  {t}^{2}}  \\  \\ \tt\purple{:  \implies 300 = 40t -  {t}^{2}}  \\  \\ \tt\green{:  \implies  {t}^{2}  - 40t  + 300 = 0} \\  \\ \tt\purple{:  \implies t =  \frac{  - ( - 40) \pm\sqrt{ { (- 40)}^{2}  - 4 \times 1 \times 300} }{2 \times 1}  }\\  \\ \tt\green{:  \implies t =  \frac{40 \pm \sqrt{1600 - 1200} }{2}  }\\  \\ \tt\purple{:  \implies t =  \frac{40 \pm 20}{2}  }\\  \\  \green{\tt:  \implies t = 10 \: sec \: and \: 30 \: sec}

7 0
3 years ago
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