Answer:
0.2129 mm
Explanation:
We have given volume of the paint = 1 gallon![=3.79\times 10^{-3}m^3](https://tex.z-dn.net/?f=%3D3.79%5Ctimes%2010%5E%7B-3%7Dm%5E3)
Area that covers the paint ![=17.8 m^2](https://tex.z-dn.net/?f=%3D17.8%20m%5E2)
We have to find the thickness of the fresh paint
So
So the thickness of fresh paint on the wall is 0.2129 mm
Answer:
The angle through which the wheel turned is 947.7 rad.
Explanation:
initial angular velocity,
= 33.3 rad/s
angular acceleration, α = 2.15 rad/s²
final angular velocity,
= 72 rad/s
angle the wheel turned, θ = ?
The angle through which the wheel turned can be calculated by applying the following kinematic equation;
![\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad](https://tex.z-dn.net/?f=%5Comega_f%5E2%20%3D%20%5Comega_i%5E2%20%2B%202%5Calpha%20%5Ctheta%5C%5C%5C%5C%5Ctheta%20%3D%20%5Cfrac%7B%5Comega_f%5E2%5C%20%20%20-%5C%20%20%5Comega_i%5E2%7D%7B2%5Calpha%20%7D%20%5C%5C%5C%5C%5Ctheta%20%3D%20%5Cfrac%7B%2872%29%5E2%5C%20%20%20-%5C%20%20%2833.3%29%5E2%7D%7B2%282.15%29%7D%5C%5C%5C%5C%5Ctheta%20%3D%20947.7%20%5C%20rad)
Therefore, the angle through which the wheel turned is 947.7 rad.
Answer:
Option C) 2,090 J/(mol K)
Explanation:
Data:
Volume in the beaker = 429 ml
temperature = 20° C
Density = 789 kg/m³
Equilibrium reading = 429
volume change = 29 ml
= 0.029 L
Energy change = mcΔT
U + PΔV
Desireurban has given a good discussion in the first answer,
but be careful with it:
-- Yes, skimming instead of reading carefully can be non-productive.
-- Using credible sources is MORE productive, and using less reliable sources
can be LESS productive.
Another nonproductive learning behavior ... one that I worry about a lot ...
is slapping the question up on Brainly and waiting for someone to give
you the answer, instead of taking some time to think about it.
Assuming the ball follows classical 2D projectile motion (moves in a parabola) and that the height y = the maximum height the ball goes in the y direction (because this would be its midpoint), then the velocity at height y is equal to the initial x component of velocity. At the midpoint, the y component is zero, so the velocity only depends on the x component. Projectiles move at constant speed in the x direction, so X = Xo. As long as you know actual values for Vi and either the initial angle or one initial component, then you can solve for Xo using trigonometry. Xo is thus the velocity of the ball once it has reached its maximum height.