Good question. The amount of 'stuff' in an object is it's mass.
A fundamental distinction we learn in physics is the difference between mass and weight. If we were in deep space, away from any very large objects of mass (like a planet), we would be 'weightless' e.g. not feel the effects of gravity, but we would not be 'massless'. Our mass doesn't change based upon our proximity to large objects (gravitational attraction), but the sense of weight does.
100cos30° + 80cos120° + 40cos233° + Dx = 0
<span>Dx = -22.53 N </span>
<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>
<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>
<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
Answer:
a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.
b) The frequency is 1404.08 Hz
Explanation:
If the police car is a stationary source, the frequency is:
(eq. 1)
fs = frequency of police car = 1200 Hz
fa = frequency of moving car as listener
v = speed of sound of air
vc = speed of moving car
If the police car is a stationary observer, the frequency is:
(eq. 2)
Now,
fL = frequecy police car receives
fs = frequency police car as observer
a) The velocity of car is from eq. 2:
b) Substitute eq. 1 in eq. 2:
