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ExtremeBDS [4]
2 years ago
11

A motorcycle has a constant acceleration of 2.5 meters per second squared

Physics
1 answer:
pshichka [43]2 years ago
7 0

Answer:

given \\ accelaration (a)= 2.5 \frac{m}{ {s}^{2} }  \\ initial \: velocity(u) = 21 \frac{m}{s}  \\ final \: velocity(v) = 31 \frac{m}{s}  \\ we \: want \: find \: time \\ newtons \: equation \: of \: motion \: v = u + at \\ 31 = 21 + 2.5 \times t \\ 31 - 21 = 2.5t \\ 10 = 2.5t \\  \frac{10}{2.5}  = t \\ t = 4sec

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On a day when the temperature reaches 50°F, the temperature in degrees Celsius is: 20°C
anzhelika [568]

Answer:

10°C

Explanation:

To convert °F to °C, we use the formula:

°C =  (°F - 32) * ( 5/9)

So, to convert 50°F to the equivalent  in °C, we can proceed as follows:  

°C = ( 50 - 32 ) * (5/9)  

°C = ( 18 ) *  (5/9), which is, approximately,

°C = 9.999999999... ≈ 10 (5/9 ≈0.555555...)

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3 years ago
What is one characteristic of the convection inside the sun?
andrew11 [14]

Answer: 3- Large cells of rising and sinking gasses

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3 years ago
Trees in the mid-west change colors and lose their leaves in the cooler parts of the year.
wariber [46]
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3 0
3 years ago
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An Argon laser (λ = 5.0×102nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the speed of
Yanka [14]

Answer:

The speed of the laser light in the cable, c_f=2.1\times 10^8\ m/s

Explanation:

It is given that,

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Let c_f is the speed of the laser light in the cable. The speed of light in a medium is given by :

c_f=\dfrac{c}{n}

c_f=\dfrac{3\times 10^8\ m/s}{1.46}

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or

c_f=2.1\times 10^8\ m/s

So, the speed of the laser light is 2.1\times 10^8\ m/s. Hence, this is the required solution.

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3 years ago
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Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h
borishaifa [10]

we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi ​ =632mi/h=632mi/h( 1mi 1609m ​ )( 3600s 1h ​ )=282m/s (a) taking v xf ​ =v xi ​ +a x ​ t with v xf ​ =0 a x ​ = t v xf ​ −v xf ​ ​ = 1.40s 0−282m/s ​ =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f ​ −x i ​ = 2 1 ​ (v xi ​ +v xf ​ )t= 2 1 ​ (282m/s+0)(1.40s)=198m

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