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Alinara [238K]
4 years ago
13

Which of the following statements is true?

Physics
2 answers:
Marianna [84]4 years ago
7 0

Answer:

D

Explanation:

all of the above

talk less smile more

If you know this qoute friend me.

seraphim [82]4 years ago
4 0
All of the above statements are true.
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You are on a school bus driving north at 20 m/s. You walk toward the back of the bus with a velocity of 6 m/s. What is your velo
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Which of the following must an engineer consider when designing a car's tires?
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I hope this answer is helpful.  The question you posted is a very difficult one, because there are no answer choices listed, and also because the question has the mysterious unknown term "the following" in it.

8 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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