D. Both exhibit the same particle-to-particle interaction.Because disturbance is propagated with the help of particles. Other than this,[ <span>light waves are electromagnetic waves. ocean waves and sound waves are mechanical waves. they are able to transfer energy. electromagnetic wave and ocean waves are transverse waves while sound waves are the longitudinal wave. they show wave properties: reflection, refraction, diffraction, interference, and plane-polarization. longitudinal waves such as sound waves cannot be plane-polarized]. The one in the box shows different examples of waves with their examples. Hope it helps.</span>
Answer:
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph

<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph

<u>3. Addition:</u>
You add the corresponding components:

To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse:
Hey there!
<span>
An object's velocity can be described by its speed and acceleration.
This statement is true
Hope this helps
Have a great day (:
</span>
Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:

d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.
Answer:
0.0195 m
Explanation:
= density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³
= diameter of hockey puck = 13 cm = 0.13 m
= height of hockey puck = 2.8 cm = 0.028 m
= density of mercury = 13.6 gcm⁻³ = 13600 kgm³
= depth of puck below surface of mercury
According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck
Weight of mercury displaced = Weight of puck
