I think this is the solution:
1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
The x -component of the object's acceleration is 2 m/s².
<h3>What's the resultant force along x- direction?</h3>
- Forces along x axis direction are as follows
- 4N along +x axis, so it's taken as +4 N
- 2N along -x axis , so it's taken as -2N.
- Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.
<h3>What's the acceleration along x axis direction?</h3>
- As per Newton's second law, Force = mass × acceleration of the object
- Force along x axis= mass × acceleration along x axis= 2N
- Acceleration = 2/ mass = 2/1 = 2 m/s²
Thus, we can conclude that the acceleration along x axis is 2 m/s².
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?
Learn more about the acceleration here:
brainly.com/question/460763
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Answer:
Net Force = 10N
Acceleration = 2m/s^2
Explanation:
calculate the net force and the acceleration on the block
Net force on the block F = mass * acceleration
Net force acting in the positive direction = 4N + 6N = 10N
Mass = 5kg
According to newton's second law;
a = F/m
a = 10N/5
a = 2m/s^2
hence the acceleration on the block is 2m/s^2
Answer:
The answer is 2,416 m/s. Let's jump in.
Explanation:
We do work with the amount of energy we can transfer to objects. According to energy theory:
W = ΔE
Also as we know W = F.x
We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>
Under the force block gains;
W = F.x → 
In the second position block has both kinetic and potential energy. Following the law of conservation of energy;
W = ΔE = Kinetic energy + Potantial Energy
W = ΔE = 
Here we can find h in the triangle i draw in the picture using sine theorem;
In a triangle 
In our situation
→ 
Therefore

→ 