Answer and Explanation:
As per the question:
When the stone is thrown from the cliff top and hits the ground below eventually:
R = 
where
= initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
Now,
(a) Displacement of the stone is given by the horizontal range:
R =
where
= initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
(b) Speed just prior to the impact is given by the third equation of motion:
where
v = final velocity
(c) Time of flight is given by the second eqn of motion where the initial velocity is considered to be 0 then:
T = 
Answer
The same number of particles in a gas spread further apart than in the liquid or solid states.
Explanation:
The same number of particles in a gas spread further apart than in the liquid or solid states. The same mass takes up a bigger volume. This means the gas is less dense. Density also depends on the material.
The angles for the first-order diffraction of the shortest and longest wavelengths of visible light are 22.33 ⁰ and 49.46 ⁰ respectively.
<h3>Angle for the first order diffraction</h3>
The angle for the first order diffraction is calculated as follows;
dsinθ = mλ
sinθ = mλ/d
<h3>For shortest wavelength (λ = 380 nm)</h3>
d = 1/10,000 lines/cm
d = 1 x 10⁻⁴ cm x 10⁻² m/cm = 1 x 10⁻⁶ m/lines
sinθ = (1 x 380 x 10⁻⁹)/(1 x 10⁻⁶)
sinθ = 0.38
θ = sin⁻¹(0.38)
θ = 22.33 ⁰
<h3>For longest wavelength (λ = 760 nm)</h3>
sinθ = (1 x 760 x 10⁻⁹)/(1 x 10⁻⁶)
sinθ = 0.76
θ = sin⁻¹(0.76)
θ = 49.46 ⁰
Learn more about diffraction here: brainly.com/question/16749356
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Answer:
A
Explanation:
In section A, she is moving in constant speed because there's a flat line which indicates speed remained same as time passed. In section c, there's a flat line but look at y axis for this one, it is 0 which means there's no speed during this part. B and D have varying speeds as there is a change in y axis variables as time passes.
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
