Let 
be the direction the swimmer must swim relative to east. Then her velocity relative to the water is
The current has velocity vector (relative to the Earth)
The swimmer's resultant velocity (her velocity relative to the Earth) is then
We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:
which is approximately 41º west of north.
Answer:
option (b)
Explanation:
According to the Pascal's law
F / A = f / a
Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.
Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm
A = π R^2 = π x 100 cm^2
F = 3 tons = 3000 kgf
diameter of plunger, d = 3 cm, r = 1.5 cm
a = π x 2.25 cm^2
Use Pascal's law
3000 / π x 100 = f / π x 2.25
f = 67.5 Kgf
The kinetic energy of the small ball before the collision is
KE = (1/2) (mass) (speed)²
= (1/2) (2 kg) (1.5 m/s)
= (1 kg) (2.25 m²/s²)
= 2.25 joules.
Now is a good time to review the Law of Conservation of Energy:
Energy is never created or destroyed.
If it seems that some energy disappeared,
it actually had to go somewhere.
And if it seems like some energy magically appeared,
it actually had to come from somewhere.
The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.
Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.
