Answer:
we can conclude that mass of air that has entered the tank is 7.80 kg.
Explanation:
given data
volume = 1.85 m³
density 1 = 1.18 kg/m³
density 2 = 5.30 kg/m³
solution
we know that density is the mass present in per unit volume so that is
mass = density × volume ........................1
here density is change so put here value in equation 1 we get
mass = ( 5.30 - 1.18 ) kg/m³ × 1.85 m³
solve it we get
mass = 7.807 kg
so we can conclude that mass of air that has entered the tank is 7.80 kg.
Answer:
6 seconds
Explanation:
Just divide 3900 by 650 to get the amount of time.
In the vertical direction, we know that acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity), we know our initial velocity (vi) = 15m/s and our final velocity (vf) = 0m/s if the ball stops at its max height. Using this kinematics equation: vf^2 = vi^2 + 2a*h, where h is the max height of the ball, we can find h by solving: h = (vf^2 - vi^2)/2a, now we plug in numbers h = (0^2 - 15^2)/2(-9.8) = 11.48 m.
Now we can find the time, but let's use a trick, we know the acceleration will be the same going up and down and that it will cover the same distance. Using this logic, we can know that the ball will be going the exact same speed we threw it up at when it comes back down to our hand (provided our hand is at the same height). We can then say it will take the same amount of time to reach its peak after leaving our hand as it will take to go from its peak back down to our hand. Using this, we can just get the time it takes to get to the top of its arc and then multiple it by 2. Using d = (vf + vi)*t / 2, we can solve for t, so t = 2d/(vf+vi) = 2(11.48)/(15) = 1.53 s for the trip up, doubling it for the trip down, we get a total of 3.06 s to go up and back down to our hand.
Answer:
17.64 N
Explanation:
Draw a free body diagram (see attached). At the center of the ladder is weight pulling down. At point C, we have a normal force pushing perpendicular to the ladder. At point A, we have reaction forces in the x and y directions.
For simplicity, we can divide the normal force into x and y components. The slope of the ladder is -80/60, so the slope of the normal force is 60/80. Therefore:
Ny / Nx = 60 / 80
Ny / Nx = 3/4
Next, we'll take the sum of moments at point A:
∑τ = Iα
-Wd + Ny (60) + Nx (80) = 0
d is the horizontal distance between A and the center of the ladder. We can find it using similar triangles. From Pythagorean theorem, we know the distance between A and C is 100 cm. So:
60 / 100 = d / 60
d = 36
-(5)(9.8)(36) + Ny (60) + Nx (80) = 0
60 Ny + 80 Nx = 1764
We now have two equations and two variables. Solving:
60 (3/4 Nx) + 80 Nx = 1764
45 Nx + 80 Nx = 1764
125 Nx = 1764
Nx = 14.112
Ny = 3/4 Nx
Ny = 10.584
Using Pythagorean theorem to find N:
N = √(Nx² + Ny²)
N = 17.64
The magnitude of the normal force is 17.64 N.
Answer:
2.4 × 10^(-11) N
Explanation:
F = Gm²/r²
F = 6.67×10^(-11) × 60² / 100²
= 2.4×10^(-11) N
