You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
Answer:
40sec
Explanation:
Data
Work = 440 J
Power= 11watt
time = ?
Power = work done/time
===> time = work done/power
= 440/11
= 40sec
They can either cancel each other or add up to a resultant force with a certain direction and modulus.
Newton's second law states that F=m*a, where F is the resultant force, ie ΣF.
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:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
