Answer:
-78.4 cm
Explanation:
We can solve the problem by using the lens equation:
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
Here we have
p = 80.0 cm
q = -39.6 cm (negative, because the image is on the same side as the object , so it is a virtual image)
Substituting, we find f:
Atom's chemical properties are determined by the number of electrons in its outermost shells.
Based on the number of electrons in the shells, it decides the reactivity of atoms.
Atoms with same number of electrons in the outer shells have similar chemical properties.
If atoms have 1,2 or 3 valence electrons they react by giving out those electrons to become stable. If atoms have 6 or 7 electrons in the outer shell they react by taking in electrons to complete its outer shell.
Answer:
3.6 × 10⁵ N/C = 360 kN/C
Explanation:
Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³
q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³
The electric field due to the point charge q₂ = 5.0 nC is E₂ = kq₂/r².
So, the magnitude of the total electric field at r = 1.0 cm is
E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)
E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)
E = 9 × 10⁹(-1 × 10⁻⁵ + 5 × 10⁻⁵)
E = 9 × 10⁹(4 × 10⁻⁵)
E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C
Answer:
the answer is option a(a) 48h in advance.
Explanation:
The second stage warning known as "CYCLONE ALERT"is assused at least 48 hours.in advanced of the expected commencement of adverse weather over the coastal areas.....the third stage warning known as "CYCLONE WARNING" issued at least 24 hours in advance of the expected commencement of adverse weather over the coastal areas.please mark me brainliest
Answer:
a) R(fw) = 46.75*10⁶ (N)
b) R(rwi) = 70.125*10⁶ [N]
Explanation: See Attached file (the rectangle stands for a jet)
The diagram shows forces acting on the jet
Let R(fw) Reaction of front wheels and
R(rw) Reaction of rear wheels
Now we apply the Stevin relation, for R(fw) and a jet weight as follows
R(fw)/ 4 = 187*10⁶ / 16
Then :
R(fw) = ( 1/4) *187*10⁶ ⇒ R(fw) = 46.75*10⁶ (N)
And e do the same for the reaction on rear wheels
R(rw) / 12 = 187*10⁶ /16 ⇒ R(rw) =(3/4)*187*10⁶
R(rw) = 140,25*10⁶ [N]
The last expression is for the whole reaction, and must be devide by 2
because that force is exerted for two wheels, therefore on each of the two rear wheels the reaction will be:
R(rwi) = 70.125*10⁶ [N]