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STatiana [176]
3 years ago
5

Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m

/s2.)
Physics
1 answer:
STatiana [176]3 years ago
7 0

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]

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An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a
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qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s

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On a safety test course, a 1000 kg car heading north at 5 m/s collides head-on with an 800 kg car heading south at 4 m/s. At the
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Answer:

The speed is 3.5\frac{m}{s} and the direction is heading north.

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In collisions the force exerted by the objects that collide is higher enough than the external forces that we can neglect that external forces, with that assumption we can use the conservation fo momentum law that states, final total momentum (pf) is equal initial total momentum (pi) if there’re not external forces or they are small enough to be neglected. Mathematically:

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m_nv_{nf}+m_sv_{sf}=m_nv_{ni}+m_sv_{si} (1)

It's important to note that when we talk about momentum and velocity direction matters, so we're are going to choose a system of reference where quantities pointing north are positive and pointing south are negative. So, the initial velocity of 1000 kg car is vni=5 m/s, initial velocity of 800 kg car is vsi=-4 m/s and the final velocity of 1000 kg car is vnf=-1 m/s. Now we can solve (1) for vsf and use the values we already have:

v_{sf}=\frac{m_nv_{ni}+m_sv_{si}-m_nv_{nf}}{m_s}=\frac{(1000)(5)+(800)(-4)-(1000)(-1)}{800}

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