Answer:
R=m*g-∀fl*g*l3
Explanation:
<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>
From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced
density of iron =mass/ volume
rho=m/l3
mass=rhol3
weight fluid=rhofluid*g*Volume
weight of fluid=rhofluid*g*l3
F=∀fl*g*l3
Downward force is weight of iron
w=m*g
Reading on the spring scale
R=w-F
R=m*g-∀fl*g*l3
m=mass of iron
g=acceleration due to ravity
rhfld=density of fluid
l3=volume of fluid displaced
Answer:
Option B.
Explanation:
Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).
Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
Answer:
(a) 3107.98 J
(b) 14530.6 J
Explanation:
mass, m = 3.56 kg
angular speed, ω = 179 rad/s
Moment of inertia of solid cylinder, I = 1/2 mr^2
where, m is the mass and r be the radius of the cylinder.
(a) radius, r = 0.330 m
I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2
The formula for the rotational kinetic energy is given by
K = 0.5 x 0.194 x 179 x 179 = 3107.98 J
(b) radius, r = 0.714 m
I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2
The formula for the rotational kinetic energy is given by
K = 0.5 x 0.907 x 179 x 179 = 14530.6 J
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
