The uniform movement allows finding the results for the helicopter displacement during the 10 s is: 700 m
Kinematics studies the motion of bodies, finding relationships between the position, velocity and acceleration of the body, in the special case that the acceleration is zero, it is called uniform motion and is described by the equation
Where v is the average velocity, Ds is the displacement and t is the time
Δs = v t
They indicate that the average speed is 70 m / s and the travel time is t=10.0s
Let's calculate
Δs = 70 10.0
Δs = 700 m / s
In conclusion using the uniform movement we can find the results for the helicopter displacement is: 700 m
Learn more here: brainly.com/question/14355103
The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)
<span>The direction of the electric field's vibration</span>
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
The reasoning for this is false
