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3 years ago

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A train car with mass m1 = 515 kg is moving to the right with a speed of v1 = 7.5 m/s and collides with a second train car. The

two cars latch together during the collision and then move off to the right at vf = 4.8 m/s.Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v1 = 7.5 m/s and collides with the second train car that is now moving to the left with a velocity v2 = -6 m/s before the collision. The two cars latch together at impact.What is the final velocity of the two-car system? (A positive velocity means the two train cars move to the right – a negative velocity means the two train cars move to the left.)

1 answer:

Anna [14]3 years ago

6 0

Answer:

For the first situation, we first need to find the mass of the second train car.

In order to do that, we apply the conservation of the amount of movement:

$515*7.5+m2*0=(m1+m2)*4.8$

and we have as a result:

m2 = 289.6875

For the second situation, also we will apply the conservation of the amount of movement:

$515*7.5-289.6875*6 = (515+289.6875)*V$

and we have as a result:

V = 2.64 (it is moving to the right)

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Answer:

a = 1 m/s²

Explanation:

given,

mass of the person = 60 Kg

Net External force exerted = 90 N

force of friction opposing the motion = 30 N

acceleration of the box = ?

Net force = External force applied - frictional force

Net force = 90 - 30

net force = 60 N

we know

F = mass x acceleration

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a = 1 m/s²

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3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

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Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

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Answer:

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