mass of iron block given as
density of iron block is
now the volume of the iron piece is given as
Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as
here we know that
= density of liquid = 916 kg/m^3
Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium
So reading of spring balance will be 16.45 N
Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block
So the other scale will read 36.47 N
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
Answer:
D. 2.8 × 10⁹ N
Explanation:
The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.
Fe= k Q₁Q₂/r²
Q₁= -0.0045 C
Q₂= -0.0025 C
r= 0.0060 m
k= 9.00 × 10 ⁹ Nm²/C²
Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²
=2.8 × 10⁹ N
The gravitational force between two object depends on their masses and on their distance.
Since the formula is
If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.
So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.
So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
