Answer:
ρ = 1.13 10⁴ km/m³
Explanation:
For this exercise we use Newton's equilibrium equation
B –W + W_scale = 0
Where B is the thrust and W_scale is the balance reading
The push is given by Archimedes' law
B = ρ_water g V
B = W- W_scale
B = m g - m_scale g
Let's calculate
B = 14.7 9.8 - 13.4 9.8
B = 12.74 N
ρ_water g V = 12.74
V = 12.74 / ρ_water g
V = 12.74 / 1000 9.8
V = 0.0013 m³
Let's use density
ρ = m / V
We replace
ρ = 14.7 / 0.0013
ρ = 1.13 10⁴ km/m³
No, it is not possible.
A compound is a substance or material constituting of two or more elements that have been chemically combined together to form a new, different substance
Any elements that have been joined together chemically can only be separated back into their constituent elements by chemical means because the bonds holding them together can only be broken using chemical means.
A good example is sodium chloride, table salt. Poisonous chlorine gas and toxic sodium metal react together whereby sodium loses one electron which chlorine readily accepts and in the process an ionic bond is formed between the two resulting in a totally new, harmless compound , sodium chloride.
Only through electrolysis can sodium chloride be separated back into sodium and chlorine gas. No physical means can be used to do that.
Answer:
Explanation:
It is given that,
Diameter of cylinder, d = 6.6 cm
Radius of cylinder, r = 3.3 cm = 0.033 m
Acceleration of the string, 
Displacement, d = 1.3 m
The angular acceleration is given by :
The angular displacement is given by :
Using the third equation of rotational kinematics as :
Here, 
Since, 1 rad/s = 9.54 rpm
So,
So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
