Answer:
Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve.<em> </em><em>Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner</em>.
The acceleration is not aligned with the direction of travel because<em> the change in velocity is at a tangent (directed away) to the direction of travel of the bus.</em>
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where 
=Angular frequency
A=Amplitude
Using the formula
Hence, the amplitude of oscillation=A=0.199
km x h = km/h
First trial: 6 x 1 = 6km/h
Second trial: 9 x 2 = 18km/h
6 + 18 = <u>24km/h</u> (Total)
Or
6 + 9 = 15 km
2 + 1 = 3h
15 + 3 = 18
15 x 2 = 30
3 x 2 = 6
30 - 6 = <u>24km/h</u>
Answer: A. Greater than 384 Hz
Explanation:
The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.
Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.
Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.
So option a is correct. i.e. now frequency will be greater than 384 Hz.
Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m
