Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa
Answer:
This question is incomplete, the options are:
A) The object absorbs most white light and refracts most green light.
B) The object refracts most white light and absorbs most green light.
C) More green light is absorbed while more red and blue light is reflected.
D) More green light is reflected while more red and blue light is absorbed.
The answer is D.
Explanation:
Light is an electromagnetic wave that contains different colours at different wavelength. The colour of light that is seen depends on the wavelength of light that is REFLECTED, while other wavelengths of light are ABSORBED. This feature is dependent on the properties of each object that received the sunlight.
For example, an object will appear GREEN because it has properties that enables it to REFLECT most of the GREEN LIGHT but absorbs most of the RED AND BLUE LIGHT in the sunlight passing through the object.
Answer:
Position of object is;
s(t) = 4t³/3 + 3t + 1
Explanation:
We are told that the velocity has an expression;
v(t) = 3.00 m/s + ( 4.00 m/s³)t²
Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;
s(t) = ∫3 + 4t²
s(t) = 3t + 4t³/3 + c
Now, we were told that at x = 1.00 m, time t = 0.000 s
Thus, plugging the values in;
1 = 3(0) + 4(0³/3) + c
c = 1
Thus,the expression for the position of the object is;
s(t) = 4t³/3 + 3t + 1
We would like to know the average net force. Therefore, we can calculate the acceleration using:
x = 0.5*a*t^2
v = a*t
where x=3.6m and v=185 m/s.
Therefore,
t=v/a and hence x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a
so a= (0.5*v^2)/x
Since we know the values for v and x, as written above, we can calculate a by plugging the numbers in.
The average net force will then be:
F = m*a,
where m=7.5kg.
Answer:
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is 
.
but once crate starts moving the force of friction is reduced .
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie .
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
