Answer:
a) 6.4 x 10^-12 cm^3
b) 17 x 10^-6 mm^2
Explanation
a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3
3
4
πr
3
V = \frac{4}{3}*3.142* 1.15^3
3
4
∗3.142∗1.15
3
= 6.3715 μm^3
1 μm^3 = 10^-12 cm^3
6.3715 μm^3 = 6.3715 x 10^-12 cm^3
==> 6.4 x 10^-12 cm^3
Answer:
The peak value of the electric field is 489.64 V/m
Explanation:
Given;
power of the laser, P = 1.0 mW = 1 x 10⁻³ W
Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m
Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²
The average intensity of the light = P / A
The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)
The average intensity of the light = 318.27 W/m²
The peak value of the electric field is given by;
Therefore, the peak value of the electric field is 489.64 V/m.
Deposition:
- when a gas changes directly to a solid
- latent heat is released
- physical change, NOT a chemical change
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
