All categories

3 years ago

Heat from burning fuel warms the walls of the firebox section of the furnace in

Physics

1 answer:

nydimaria [60]3 years ago

7 0

Heat from burning fuel warms the walls of the firebox section of the furnace in

A. a hot-water heating system.

B. a hot-air heating system.

C. a compressor compartment.

D. an evaporation system.

You might be interested in

Un ascensor sube con una velocidad constante de 2 m/s. Un niño que va en el ascensor lanza una piedra de 0.6 kg hacia arriba con

Natasha_Volkova [10]

Answer:

W = 7.5 J

Explanation:

W = KE

W = ½mv² = ½0.6(5²) = 7.5 J

Niño, piedra y ascensor están todos en el mismo marco de referencia inercial.

los 2 m / s no son importantes.

6 0

3 years ago

Why e=mc2?why not e=mc3?

Softa [21]

For starters, this question isn’t really about relativity. It’s about energy, and E=mc^2 only makes sense if energy has the units of (mass)*(velocity)^2. So we might as well ask: why is kinetic energy defined as KE = ½*mv^2?

4 0

4 years ago

What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.

Luden [163]

Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.

8 0

4 years ago

A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o

irina [24]

Answer

a = 3.674 m / s ^ 2

t = 3.13 s

Using the kinematic equations for the movement we have:

$h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2$ (1)

$V_{f} = V_{0} + at$ (2)

Where:

$P_{0}$ = initial position

$V_{0}}$ = initial velocity

a = acceleration

t = time in seconds

$V_{f}$ = final speed

We know:

$P_{0}=0$

$V_{0}= 0$

h = 18 m

$V_{f} = 11.5\frac{m}{s}$

So:

From (2) we have that: $t =\frac{V_{f}}{a}$

$t =\frac{11.5}{a}$

From (1) we have to:

$h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2$

Then we clear "a" to find the acceleration.

$\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2$

Then, the time it takes to reach this speed is:

$t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s$

4 0

3 years ago

Read 2 more answers

An ice cube of mass 50.0 gg can slide without friction up and down a 25.0 degreedegree slope. The ice cube is pressed against a

kompoz [17]

<em>We have assumed the distance to be 0.1 meters (not millimeters) since the question has issues when expressing the units</em>

Answer:

$d=0.60\ m$

Explanation:

<u>Energy Conversion</u>

We need to understand and apply the concepts of energy conversion to solve this problem. Three types of energy are manifested in the motion of the ice cube of mass m.

When it's above the ground level at a height h, it has potential gravitational energy, given by

$U=mgh$

If the cube is moving at speed v, it has kinetic energy, given by

$\displaystyle K=\frac{1}{2}mv^2$

Finally, when it compresses the spring, it has elastic energy:

$\displaystyle E=\frac{1}{2}kx^2$

Where x is the distance of compression and k is the spring constant

When the ice cube is released, it has potential gravitational energy which magnitude we cannot calculate since we don't have the height. Then it goes down the slope and acquires speed and kinetic energy until it stops when compressing the spring a distance x. In that very moment, the total energy is stored in its elastic form:

$\displaystyle E=\frac{1}{2}25\cdot 0.1^2=0.125\ J$

When the ice cube travels up powered by that energy, it has both kinetic and potential energies, and it stops up in the ramp and starts reversing direction when it runs out of speed, thus the total potential energy is

$mgh=0.125\ J$