What happens when velocity and acceleration are not in the same direction?

Physics

2 answers:

hoa [83]3 years ago

6 0

I think it becomes a negative integer?

denis23 [38]3 years ago

5 0

Acceleration is the rate at which velocity changes, which means that acceleration tells you how fast the velocity is changing. A large acceleration tells you that the velocity is changing quickly - a small acceleration tells you that the velocity is changing slowly - an acceleration of zero tells you that the velocity is not changing at all.

Acceleration tells you how the velocity changes - it doesn't tell how how much the velocity is! An object can have a large velocity and a small (or zero) acceleration - and vice versa.

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2 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Rashid [163]

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Explanation:

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slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$