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kramer
3 years ago
5

A marble accelerates from rest at a constant rate .it travels 36 m in 12.0 sec what’s is it’s final velocity and what was the ac

celeration

Physics
1 answer:
ddd [48]3 years ago
4 0

Answer:

a = 0.50 m/s²

v = 6.0 m/s

Explanation:

Step 1: Given data

  • Initial velocity (u): 0 m/s (rest)
  • Displacement of the marble (s): 36 m
  • Time elapsed (t): 12.0 s
  • Final velocity (v): ?
  • Acceleration (a): ?

Step 2: Calculate the acceleration

We will use the following suvat equation.

s = u × t + 1/2 × a × t²

s = 0 × 12.0 s + 1/2 × a × (12.0 s)²

36 m = a × 72.0 s²

a = 0.50 m/s²

Step 3: Calculate the final velocity

We will use the following suvat equation.

v = u + a × t

v = 0 + 0.50 m/s² × 12.0 s

v = 6.0 m/s

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please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c
-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, A_x=A\text{Sin}37

A_x=12[\text{Sin}(37)]

     = 7.22 m

Vertical component, A_y=A[\text{Cos}(37)]

    = 9.58 m

Similarly, Horizontal component of vector C,

C_x  = C[Cos(60)]

     = 6[Cos(60)]

     = \frac{6}{2}

     = 3 m

C_y=6[\text{Sin}(60)]

    = 5.20 m

Resultant Horizontal component of the vectors A + C,

R_x=7.22-3=4.22 m

R_y=9.58-5.20 = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

R=\sqrt{(R_x)^{2}+(R_y)^2}

   = \sqrt{(4.22)^2+(4.38)^2}

   = \sqrt{17.81+19.18}

   = 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = \frac{\text{CB}}{\text{OB}}

                  = \frac{R_y}{R_x}

                  = \frac{4.38}{4.22}

m∠COB = \text{tan}^{-1}(1.04)

             = 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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3 years ago
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