Cavity wall insulation costs £240 but will save you £32 each year. What is the payback time for cavity

Physics

1 answer:

likoan [24]2 years ago

5 0

Answer:

Thus, the payback time for cavity wall insulation is 7.5 years

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To study waves that humans can see which section of the electromagnetic spectrum would you use

andrey2020 [161]

Visible light is the spectrum that humans can see ranging from 400-700 nm

4 0

3 years ago

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

3 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

zmey [24]

Answer:

+0.231 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum. In fact, we have that the total momentum before the collision must be equal to the total momentum after the collision:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$