D
Giddy UP!!!!!!!!!!!!!!!!!!!!!
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P =
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
Answer:
1050 kg
Explanation:
The formula for kinetic energy is:
KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.
We can now substitute the values we know into this equation.
KE = 472 500 J and v = 30 m/s:
472 500 = 1/2 × m × 30²
Next, we can rearrange the equation to make m the subject and solve for m:
m = 472 500 ÷ (1/2 × 30²)
m = 472 500 ÷ 450
m = 1050 kg
Hope this helps!
Answer:
The true course: north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
- = velocity of wind =
- = velocity of plane in still air =
Assume:
- = resultant velocity of the plane
- = direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.
Let us find the direction of this resultant velocity with respect to east direction:
This means the the true course of the plane is in the direction of north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.
Hence, the ground speed of the plane is 96.68 km/h.