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3 years ago

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

✤ Final velocity of mobile = 115 m/s

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Read 2 more answers

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

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In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

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2 years ago

What is the pendulum length whose period is 2.0s ?

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$Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m$ $T=2 \pi \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi } = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
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$

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