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marshall27 [118]
3 years ago
10

The burning of wood in a campfire is which type of reaction?

Physics
1 answer:
grin007 [14]3 years ago
4 0
The answer is C
Hope this help
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A 2 N and an 8 N force pull on an object to the right and a 4 N force pulls on the same object to the left. If the object has a
Elden [556K]

Answer:

a = 12 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF = sum of forces acting on a body [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

Let's take the direction of positive forces to the right and negative forces directed to the left

2 + 8 - 4 = 0.5*a

6 = 0.5*a

a = 12 [m/s²]

7 0
3 years ago
I need to fill the gaps
andriy [413]

Answer:

1. combustion

2. oxygen

3. CO²

5 0
3 years ago
A book is on the table. If the weight of the book is 25 newtons, what is the magnitude and direction of the normal force?
Andrej [43]
I'm not sure about the magnitude, but the direction of the normal force is upward. 
4 0
3 years ago
Read 2 more answers
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

4 0
3 years ago
Cuales son las carreras de velocidad de medio fondo y fondo?
sammy [17]

Answer:

57490

Explanation:

58=8990

6 0
3 years ago
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