A pile driver is raised to a height if 3. 0m. How high would another pile driver with twice the mass of the first have ti be raised in order to have the same amount of potential energy? Please draw the work out! (20 points!)

Answers

Answer 1

The second pile driver must be raised to a height of 1.5m.

Assume the mass of the first pile driver is m and its height is h. Therefore, the potential energy (PE) of the first pile driver is given by:

PE1 = m * g * h

where g is the acceleration due to gravity.

Now, let's find the potential energy of the second pile driver, which has twice the mass of the first pile driver. The mass of the second pile driver is 2m.

To have the same amount of potential energy as the first pile driver, the second pile driver must be raised to a certain height, let's call it h2.

Therefore, the potential energy (PE2) of the second pile driver is given by:

PE2 = (2m) * g * h2

Since we want the potential energy of both pile drivers to be equal, we can set up an equation:

PE1 = PE2

m * g * h = (2m) * g * h2

We can cancel out the mass and acceleration due to gravity:

h = 2 * h2

Now we can solve for h2:

h2 = h / 2

Plugging in the value of h as 3.0m, we have:

h2 = 3.0m / 2

h2 = 1.5m

Therefore, the second pile driver, with twice the mass of the first pile driver, must be raised to a height of 1.5m in order to have the same amount of potential energy.

Here's a visual representation of the work:

First pile driver:

Potential energy (PE1) = m * g * h

Second pile driver:

Potential energy (PE2) = (2m) * g * h2

Since PE1 = PE2, we have m * g * h = (2m) * g * h2

Cancelling out mass and acceleration due to gravity, we get h = 2 * h2

Solving for h2, we find h2 = h / 2

Plugging in the value of h, we have

h2 = 3.0m / 2

    = 1.5m

Therefore, the second pile driver must be raised to a height of 1.5m.

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Related Questions

What is the maximum speed of a point on the outside of the wheel 15 cm from the axle?.

Answers

The maximum speed of a point on the outside of the wheel 15 cm from the axle would depend on the rotational speed of the wheel.

To calculate the maximum speed, we need to know the angular velocity of the wheel, which is the rate at which it rotates. If we assume that the wheel is rotating at a constant angular velocity, we can use the formula v = rω, where v is the linear velocity of the point on the outside of the wheel, r is the radius of the wheel (15 cm in this case), and ω is the angular velocity of the wheel in radians per second.

So, if we know the angular velocity of the wheel, we can plug it into this formula and calculate the maximum speed of a point on the outside of the wheel 15 cm from the axle.

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Show that 1Kwh is equal to 3.6MJ of energy.​

Answers

Answer:

3.6 MJ

Explanation:

1 kWh = 1 MJ

Remember that this is the same as the equation Power×time = Energy

Step 1: Convert kWh (kiloWatt×hour) to Ws (Watt×second)

1 kW = 1000 Watt

1 h = 60 min×60 sec = 3600 seconds

1000 W×3600s = 3600000 Joules

Divide 3600000 J by 10^6 to get 3.6 Mega Joules

The maximum allowable resistance for an underwater cable is one hundredth of an ohm per
meter and the resistivity of copper is 1. 54 x 10-80m.
a) Calculate the smallest cross sectional area of copper cable that could be used. ​

Answers

The copper cable's smallest possible cross-sectional area is 1.54 x 10-6 square meters.

To calculate the smallest cross-sectional area of the copper cable, we can use the formula for resistance:

R = ρ(L/A),

where R is the resistance (in ohms), ρ is the resistivity of the material (in ohm meters), L is the length of the conductor (in meters), and A is the cross-sectional area (in square meters).

Given the maximum allowable resistance (R) is 0.01 ohms per meter (one-hundredth of an ohm per meter) and the resistivity of copper (ρ) is 1.54 x 10^-8 ohm meters. Let's calculate the smallest cross-sectional area (A) that can be used.

First, we'll rewrite the formula for A:

A = ρ(L/R).

Since R is given as ohms per meter, we can set L to 1 meter for simplicity, and the formula becomes:

A = ρ(1/R).

Now, we can plug in the given values:

A = (1.54 x 10^-8)/(0.01).

A = 1.54 x 10^-6 square meters.

So, the smallest cross-sectional area of the copper cable that could be used is 1.54 x 10^-6 square meters.

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