Answer:
V = 0.05 m^3
Explanation:
The weight of the object of mass (m) and volume (V) in water is: 275 N and it is calculated using the buoyancy formula:
275 N = m g - V (dw) g
where (dw) is the density of water (1000 kg/m^3)
The weight of the object in oil is 325 N and it is calculated using the buoyancy formula as:
325 N = m g - V (do) g
where "do" is the density of oil which is: 0.9 x 1000 kg/m^3 = 900 kg/m^3
Therefore, subtracting term by term the two equations:
325 N -275 N = mg - mg - V (do) g + V (dw) g
50 N = V g (dw - do)
50 N = V g (1000 - 900)
50 N = V g (100)
solving for V (which will result in units of m^3):
V = 50 / 981 m^3
V = 0.05 m^3
Answer: B. irregular galaxy
Explanation: "An irregular galaxy is a galaxy that does not have a distinct regular shape, unlike a spiral or an elliptical galaxy."
Answer:
Ring v² = gh
solid wheel (cylinder) v² = 4/3 gh
Explanation:
Let's use conservation of energy to find the speed of the wheels at the bottom of the hill.
starting point. Point before starting movement
Em₀ = mgh
final point. At the bottom of the hill
Em_f = K = ½ m v² + ½ I w²
energy is conserved
Emo = Em_f
mgh = ½ m v² + ½ I w²
angular and linear velocity are related
v = w r
we substitute
mgh = ½ m v² + ½ I v² / r²
mgh = ½ (m + I / r²) v²
v² =
the moments of inertia are tabulated
Ring
I = mr²
v² = 2 m g h / (m + m)
v² = gh
solid wheel (cylinder)
I = ½ m r²
v² = 2m gh / (m + m / 2)
v² = 4/3 gh
We can see that due to the difference in the moment of inertia of each body it is different, the solid wheel has more speed when it reaches the lower part of the ramp
The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine