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svetoff [14.1K]
2 years ago
14

An 6 kg object accelerating from 17 m/s to 10 m/s. What is the change in momentum of the object?

Physics
1 answer:
blagie [28]2 years ago
4 0

Answer:

The change in momentum (in kg*m/s) = 42

Explanation:

The change in momentum is equal to

Final Momentum - Initial Momentum

As we know

Momentum is equal to the product of mass and velocity

M = m * v

where m is the mass of the body and v is the speed

Thus, change in momentum

m_2v_2 - m_1v_1

here m1 = m2 = 6 Kg

Substituting the given values in above equation we get -

6 * 17 - 6* 10\\6 (17-10) = 6 * 7 = 42  kg 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
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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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0.01 kg has a weight 0.081 Newtons

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