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disa [49]
3 years ago
6

A car moving west slows down from a velocity of 50 m/s to 20 m/s in 15 seconds. What is its acceleration?

Physics
1 answer:
velikii [3]3 years ago
6 0
Vf = Vo + at

Vf = 20 m/s
Vo = 50 m/s
a = ? 
t = 15

Therefore

20 = 50 + 15a

20 - 50 = 15a

-30 = 15a

a = -30 / 15

a = -2 m/s²
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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?
Korvikt [17]

\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>

4 0
3 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

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