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Fed [463]
3 years ago
7

A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?

Physics
1 answer:
Korvikt [17]3 years ago
4 0

\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

  • u = 250 m/s
  • a = -3 m/s²
  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

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A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

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The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is
dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

\lambda = \frac{x*d}{L}        

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots                      </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

\lambda_{1} = \frac{x_{1}*d}{L}

630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}

\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}  (1)    

For the second wavelength of light we have:

\lambda_{2} = \frac{x_{2}*d}{L}

\lambda_{2} = 0.45 m*\frac{d}{L}   (2)  

By entering equation (1) into equation (2) we have:

\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm

Therefore, the second wavelength is 464.8 nm

I hope it helps you!          

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