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3 years ago

Carbon and oxygen react to form carbon dioxide what are the products?

Physics

2 answers:

jeyben [28]3 years ago

5 0

1 carbon and 2 oxygen

Harman [31]3 years ago

4 0

Answer:

When a non-metal burns in oxygen, a non-metal oxide forms as product. Here is the word ... Coal is a form of carbon and when it burns in oxygen we can represent the reaction with the following word equation: carbon + ... Which non-metal do you think reacts with oxygen to form sulfur dioxide? See if you ... C + O2 → CO2.

Explanation:

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Match the chemical formula to the correct name.

11Alexandr11 [23.1K]

Co carbon monoxide
sorry friend i don't know other ones

3 0

3 years ago

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

Some one help my science homework is due tomorrow and I'm so stuck with question 8-9, and 11-12

yuradex [85]

Off the top of my head, I only know 9 and 11, so I'll answer those two.

9) A heterotroph is an organism that relies on other organisms for food/energy
An autotroph can produce its own food from inorganic compounds (light)

11) Vascular plants have specialized tubes for transporting nutrients
Nonvascular plants do not have such tubes and are simpler

5 0

3 years ago

A pitcher throws a curveball that reaches the catcher in 0.61 s. The ball curves because it is spinning at an average angular ve

sashaice [31]

Answer:

22.36 rad

Explanation:

Applying,

ω = θ/t.............. Equation 1

Where ω = angular velocity, θ = angular displacement of the baseball, t = time

make θ the subject of the equation

θ = ωt............... Equation 2

From the question,

Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s

Substitute these values into equation 1

θ = 0.61(36.652)

θ = 22.36 rad

Hence the angular displacement of the baseball is 22.36 rad

4 0

3 years ago