The volume of the object must be no larger than 
.
Explanation:
In order for an object to be able to float in water, its density must be equal or smaller than the water density.
The density of water is:
This means that the density of the object must be no larger than this value.
We also know that the density of an object is given by
where
m is the mass of the object
V is its volume
For the object in this problem, the mass is
Therefore, we can re-arrange the equation to find its volume:
So, the volume of the object must be no larger than .
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It is number 3 because I know it is
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0
Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)
Answer:
Explanation:
Let's answer these statements
.1) True. This is the law of reflection.
.2) False. The speed of light depends on the index of refraction n = c / v
v = c / n
.3) True. The frequency creates a forced oscillation, whereby the atoms re-emit at the same incident frequency
.4) False. The index of refraction is a measure of the ratio of the speed of light in a vacuum and the material environment, the ability to change the trajectory is given by the law of refraction
.5) True. True due to the change in beam trajectory due to the law of refraction
.6 False. The phenomenon occurs when you pass from a medium with a higher index to one with a lower ratio, because the refracted beam separates from the normal
.7) True.
.8) False so that the lightning approach is valid Lam >> d,
.9) True.
Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is 
, and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
