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3 years ago

A particle is constrained to move along a straight line through O.

1 answer:

7 0

x=1.9+2.5t+3.7t2 V=2.5−3.7t V=0 when it stops 0=2.5-7.4t t=0.3378 a) x=1.9+2.5t+3.7t2 (t=0.3378) x=3.1167m b) V=2.5−3.7t (t=0) V=2.5 but because it's returning here velocity is -2.5

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3 years ago

Read 2 more answers

Add the vectors:

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Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

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3 years ago

What is the main reason for using a data table for collecting data?

Luba_88 [7]

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3 years ago

The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t

DiKsa [7]

I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
$E_p=\frac{1}{2}kx^2$
The kinetic energy of two carts is:
$E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
So we have:
$E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
Momentum also has to be conserved:
$m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2$
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
$v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}$
Now we go back to the momentum equation:
$v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}$

8 0

2 years ago