1 Kilojoule [kJ] = 737.562 149 277 27 Foot pound force [ftlbf]
Answer:
Explanation:
Mass = 624 gm = .624 kg
weight = .624 x 9.8
= 6.11 N
Radius of ball = 12.15 x 10⁻² cm
volume of ball
= 4/3 x 3.14 x ( 12.15 x 10⁻²)³
= 7509.26 x 10⁻⁶ m³
Buoyant force = weight of displaced water
= 7509.26 x 10⁻⁶ x 10³ x 9.8
= 73.59 N
b ) Since buoyant force exceeds the weight of the ball , it will float .
c )
Let volume v sticks out while floating .
Volume under water
= 7509.26 x 10⁻⁶ - v
its weight
= (7509.26 x 10⁻⁶ - v ) x 10³ x 9.8
For floating
(7509.26 x 10⁻⁶ - v ) x 10³ x 9.8 = .624 x 9.8 ( weight of ball )
(7509.26 x 10⁻⁶ - v ) x 10³ = .624
7.509 - v x 10³ = .624
v x 10³ = 7.509 - .624
v x 10³ = 6.885
v = 6.885 x 10⁻³ m³
fraction
= v / total volume
= 6.885 x 10⁻³ / 7.51 x 10⁻³
91.67 %
<span>Rays will follow the law of reflection, so the angle of reflection will be 35 degrees. If a</span> light ray strikes a smooth surface, the reflected ray will bounce off the surface with the same angle the ray hits the surface. In other words, the angle of incidence is the same of angle of reflection, which is 35 degrees. If the surface is not smooth, the reflected ray might diffuse in all directions.
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = - 1.4 m
