The particles always move parallel and perpendicular to the waves. The waves which are in the water moves a circle. Both up and down and back and forth.
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<span>Yes, the vast majority of an atom is indeed empty space. Most of it's mass is centered in the nucleus. Flying around the nucleus are the electrons, but they're very very far away (on an atomic level anyway). Most of the atom is the space between the nucleus and the electrons.</span>
Given Information:
Frequency of horn = f₀ = 440 Hz
Speed of sound = v = 330 m/s
Speed of bus = v₀ = 20 m/s
Answer:
Case 1. When the bus is crossing the student = 440 Hz
Case 2. When the bus is approaching the student = 414.9 Hz
Case 3. When the bus is moving away from the student = 468.4 Hz
Explanation:
There are 3 cases in this scenario:
Case 1. When the bus is crossing the student
Case 2. When the bus is approaching the student
Case 3. When the bus is moving away from the student
Let us explore each case:
Case 1. When the bus is crossing the student:
Student will hear the same frequency emitted by the horn that is 440 Hz.
f = 440 Hz
Case 2. When the bus is approaching the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330+20 )
f = 440 ( 330/ 350 )
f = 440 ( 0.943 )
f = 414.9 Hz
Case 3. When the bus is moving away from the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330-20 )
f = 440 ( 330/ 310 )
f = 440 ( 1.0645 )
f = 468.4 Hz
Answer:
D. 0.9
Explanation:
Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.
Resolving into x component:
∑
------(1)
Resolving into y component:
∑
------(2)
Static frictional force,
μ
------(3)
substituting
from equation (1) and
from equation (2) into equation (3)
μ
μ
μ 
μ 
The angle the vehicles make with the horizontal α = 42°
μ ≥ tan 42°
μ ≥ 0.9
Answer:
3.99*10^-3N/C
Explanation:
Using
Ep= kq/r²
Where r = 0.6mm = 0.6*10^-3m
K= 8.9*10^9 and q= 1.6*10^-19
So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²
= 3.99*10^-3N/C