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3 years ago

In addition to ash, what is produced when wood burns

Physics

1 answer:

Genrish500 [490]3 years ago

8 0

I believe the answer is: B (oxygen and carbon dioxide)

I am not for sure, but good luck!

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An unknown material has a mass

atroni [7]

Answer: 1896.55J/kg°C

Explanation:

The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 1320 joules

Mass of material = 5.61kg

C = ? (let unknown value be Z)

Φ = 0.124°C

Then, Q = MCΦ

1320J = 5.61kg x Z x 0.124°C

1320J = 0.696kg°C x Z

Z = (1320J / 0.696kg°C)

Z = 1896.55 J/kg°C

Thus, the specific heat of the material is 1896.55J/kg°C

4 0

3 years ago

Read 2 more answers

The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

$\frac{P_{1} }{\\P_2}$ = $\frac{density * g * h_1}{density * g * h_2}$

So $\frac{39}{P_2}$ = $\frac{3}{9}$

Or P₂= 39 * 3 = 117 kPa

5 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

A 20 metric ton train moves toward the south at 50 m/s. At what speed must it travel to have four times its original momentum

OverLord2011 [107]

Answer:

200 m/s

Explanation:

as momentum is a product of mass and speed, and mass is not changing, four times the speed will result in four times the momentum.

p = mv

4p = m(4v)

5 0

2 years ago

What is step one of a star called?

yKpoI14uk [10]

Answer: Stage 1- Stars are born in a region of high density Nebula, and condenses into a huge globule of gas and dust and contracts under its own gravity. This image shows the Orion Nebula or M42 . Stage 2 - A region of condensing matter will begin to heat up and start to glow forming Protostars.

Explanation:

3 0

3 years ago