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3 years ago

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A skateboarder rolls off a ledge that is 1.12 m high, and lands 1.48 m from the base of the edge. How much time was he in the ai

r?

1 answer:

nignag [31]3 years ago

3 0

The rider's horizontal motion, and how much ground he covers before he hits it, have nothing to do with how long he takes to hit the ground. The problem is simply: "How long does it take an object to fall 1.12 m from rest ?"

This seems like a good time to use this formula:

Distance fallen from rest = (1/2) (acceleration) (time)²

The problem doesn't tell us what planet the skateboarder is exercising on. I'm going to assume it's on Earth, where the acceleration of gravity is 9.8 m/s². And now, here's the solution to the problem I just invented:

1.12 m = (1/2) (9.8 m/s²) (time)²

Time² = (1.12 m) / (9.8 m/s²)

Time² = 0.1143 sec²

Time = √(0.1143 sec² )

<em>Time = 0.34 second</em>

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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

alukav5142 [94]

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: Kinetic energy is energy in motion. For example would be the energy used or the momentum used to make a car drive down a hill. The kinetic energy of an object is proportional to to the square of its speed.

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Answer:

The work done by Joel is greater than the work done by Jerry.

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Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

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Answer:

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Explanation:

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