Answer:
Density (φ) = 0,8827 Kg/L
Specific weight (Ws) = 8,65 N/L
Specific gravity (Gs) = 0,8827 (without unit)
Explanation:
The density formula: φ = 
I know the mass "m", I need to find out the volume of the cylinder (V)
V = π* r²*h
The radius "r" is equal to half the diameter (150mm) = 75mm
Now I can find out the density (φ)
φ = 
= 0,8827 Kg/L
The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:
Ws = φ*g
(g = gravity = 9,8 m/s²)
Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :
Gs = φ(o) / φ(w)
(φ(w) = 1 Kg/L
Hope this can help you !!
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer: Option (c) is the correct answer.
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
When we increase the temperature of a liquid substance then there will occur an increase kinetic energy of the molecules. As a result, they will move readily from one place to another.
Hence, liquid state of a substance will change into vapor state of the substance. This means that an increase in temperature will lead to an increase in vapor pressure of the substance.
Thus, we can conclude that you can increase the vapor pressure of a liquid by increasing temperature.
