If you are asking what the volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.
Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.
Explanation:
First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.
From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.
Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
Answer:
Explanation:
first write the newtons second law:
F
=δma
Applying bernoulli,s equation as follows:
∑
Where, 
is the pressure change across the streamline and 
is the fluid particle velocity
substitute 
for {tex]γ[/tex] and 
for 
integrating the above equation using limits 1 and 2.
there the bernoulli equation for this flow is
note: 
=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
