Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:
Answer:
Explanation:
A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell outwardly oriented and end at negatively charged shell .
B )
distance between the centres of spherical shell
= 2 a
potential energy of charges
= k q₁ x q₂ / R
= k x - Q x Q / ( 2a )
= - k Q²/ 2a
So work needed to separate them to infinity will be equal to
= k Q²/ 2a
Answer:
given,
mass of copper = 100 g
latent heat of liquid (He) = 2700 J/l
a) change in energy
Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (300 - 4)
Q = 11153.63 J
He required
Q = m L
11153.63 = m × 2700
m = 4.13 kg
b) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (78 - 4)
Q = 2788.41 J
He required
Q = m L
2788.41 = m × 2700
m = 1.033 kg
c) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (20 - 4)
Q = 602.90 J
He required
Q = m L
602.9 = m × 2700
m =0.23 kg
m₁ = mass of the first object = 3.0 kg
m₂ = mass of the second object = 3.0 kg
r = distance between the first and second object = 1.0 m
G = universal gravitational constant = 6.67 x 10⁻¹¹ N m²/kg²
F = force of gravity between the two objects = ?
according to law of gravitation, force of attraction "F" between two objects m₁ and m₂, placed distance "r" apart is given as
F = G m₁ m₂/r²
inserting the values
F = (6.67 x 10⁻¹¹) (3.0) (3.0)/(1.0)²
F = (6.67 x 10⁻¹¹) (9.0)
F = 60.03 x 10⁻¹¹ N
F = 6.003 x 10⁻¹⁰ N