1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k
2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2
Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)
For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
Is this a true or false question?
That is because work requires energy. According to the law of conservation of energy, it cannot be created or destroyed. When doing work, energy change forms and gets transferred to the object until it is released.
for example, when you lift up an object and place it on a higher elevation, you transferred energy to it and gave it potential energy. The potential energy is transformed into kinetic energy when the object falls down, and if it hits a surface, the energy will scatter, vibrating the areas around it and producing sound.
Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy
Your answer is correct. No problem and Have a nice day
Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
P = F / A
Where the area of a circle is
A = π r²
E radius is half the diameter
r = d / 2
A = π d² / 4
We replace
P = F 4 / π d²2
P₁ = 397 4 /π 0.058²
P₁ = 1.50 10⁵ Pa
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
P₁ = P₂ + ½ ρ v₂²
v₂ = √ (P1-P2) 2 / ρ
v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
v₂ = 97.4 m / s
