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lord [1]
3 years ago
15

The largest water tower in Camden holds 2,000,000-gallons at 24 meters in the air.

Physics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

Explanation:

Given that,

The Camden holds a volume of

V = 2,000,000 gallons

At a height of

h = 24 m

We want to find the potential energy.

And potential energy can be calculated using

P.E = mgh

Where

m is the mass

g Is the acceleration due to gravity

h Is the height.

Given that,

Then volume is

V = 2,000,000 gallons

From research.

1 gallon ≈ 3.785 kg

Then,

2,000,000 gallons = 2,000,000 × 3.785 kg

So,

P.E = mgh

P.E = 2,000,000 × 3.785 × 9.81 × 24

P.E = 1.782 × 10^9 J.

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An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the
hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg

7 0
2 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
Assume you push a 20 kg child in a little red wagon with a force of 100 N. However, the dirt and rocks below the wheels
GenaCL600 [577]

Answer:

a

Explanation:

4 0
3 years ago
An athlete runs a 100 m race in 10 seconds against a friction force of 100N. How do I work out his power output?
Rashid [163]
They seem to cancel each other out which is odd
6 0
3 years ago
A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
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