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4 years ago

9

What type of reaction is: coal + oxygen -> carbon dioxide + water? A. Double Replacement B. Combustion C. Single Replacement

D. Decomposition

1 answer:

vodomira [7]4 years ago

8 0

Answer:

the answer is B. Combustion

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4. The figure below shows a complex wave pattern on a string

Marat540 [252]

Explanation:Consider the wave pattern image reflected about the rigid hook on the wall. · 005(part2of3)10.0 pointsHow many complete waves are emitted in this time

5 0

3 years ago

A 50 mm diameter thin walled pipe is covered with an insulation layer with thicknessof 25mm and thermal conductivity of0.075W/mK

Wittaler [7]

Answer:

The steam will start to condense at 6.6 mm into the pipe

Explanation:

The volume flow rate =π×(50/1000)²/4×10 = 0.0196 m³/s

The specific volume of the steam = 1.769 m³/kg

Therefore;

The mass flow rate = 0.0196/1.769 = 0.011099 kg/s

The resistance of the insulation material = ln(0.075/0.05)/(2×π×0.075) = 0.860 K/W

The resistance of the outside film of the insulator = 1/(15×2×π×0.075×1) = 0.14147 K/W

The total resistance = 0.14147 + 0.860 = 1.00147 K/W

1/(UA) = 1.00147 K/W

A = 2×π×0.05×1

1/U = 0.3146

U = 3.178 W/m² K

We have;

T(x) = T₀ + (Tin - T₀) exp(-UπDx/mcp)

Therefore, when T(x) = 100°C, we have;

100 = 20 + (120 - 20)exp(-3.178×π×0.05x/(0.011099 × 1.33))

Solving, we get

x = 6.597× 10⁻³ m ≈ 6.6 mm

Therefore, the steam will start to condense at 10 mm into the pipe.

3 0

4 years ago

At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of

photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

6 0

3 years ago

Read 2 more answers

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g

MissTica

The Earth's radius is 6371 km. So that's our distance from the center when we're on the surface.

The Shuttle astronaut's distance from the center, when s/he's in orbit, is 330 km greater ... that's 6701 km.

The force of gravity is inversely proportional to the distance between the center of the Earth and the center of the astronaut. So, in orbit, it's

(6371/6701)^2 = 90.4 %

of its value on the surface.

8 0

3 years ago

For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

Scilla [17]

Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.

The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.

The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, $\frac{dD}{d \theta} =0$
That is,
$\frac{2V^{2}}{g} cos(2 \theta )=0$

Because $\frac{2V^{2}}{g} \neq 0$ , therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.

It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.

4 0

4 years ago