Answer:
1-d
2- a
3-e
4-b
5-c
This is the correct sequence
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?
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False because currents do not flow easily through insulators. If it only said conductors, then it would be true.
Hello!
First one we can use that PE=mgh so we have
4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m
Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have
g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2
Hope this helps. Any questions please ask. Thank you.
Answer: The height of the fluid rise is 0.01m
Explanation:
Using the equation
h = (2TcosѲ )/rpg
h= height of the fluid rise
diameter of the tube =3mm
radius of the tube= 3/2 =1.5mm=0.0015
T= surface tension = 600mN/m=0.6N/m
Ѳ = contact angle =
C
p= density =3.7g/cm3= 3700kg/m3
g= acceleration due to gravity =9.8m/s2
h = ( 2*0.6*0.5)/(0.0015*3700*9.8)
h = 0.6/54.39
h= 0.01m
Therefore,the height of the fluid rise is 0.01m