Answer:
30 cm
Explanation:
To solve this problem, we use the lens equation:

where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we know that
q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image
We also know that the size of the image is twice that of the object, so the magnification is 2:

where M is the magnification of the lens. Solving this equation for p,

So, the distance of the object from the lens is 15 cm.
Now we can finally solve the lens equation to find f, the focal length:

<span>mass and only mass ................</span>
Moving point charges, such as electrons, produce complicated but well known magnetic fields that depend on the charge, velocity, and acceleration of the particles. Magnetic field lines form in concentric circles around a cylindrical current-carrying conductor, such as a length of
Jackson sells lemonade for $2 per cup. he paid $20 for supplies to sell 5 cups. What is the domain for this scenario
Answer:
W =23807.68 N
Explanation:
given,
surface area of wing = 19.4 m²
speed over top wing = 67 m/s
speed under wing = 51 m/s
density of air = 1.3 kg/m³
weight of plane
From Bernoulli's principle

where 1 and 2 are two different locations at the same geo potential level
so if we call 1 the lower surface and 2 the upper surface,
we find the pressure differential, P₁ -P₂
then the force acting on the plane is
F=P A
F=1227.2 x 19.4
F =23807.68 N
weight of the plane
W =23807.68 N