<span>(a)
Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:
18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.
(b)
To much the same end do we derive the vertical component:
18.0*sin37.5 = v_y = 10.96 ms^-1
Which we then divide by acceleration, a_y, to derive the time till maximal displacement,
10.96/9.8 = 1.12 s
Finally, doubling this value should yield the particle's total time with r_y > 0
<span>2.24 s
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span></span>
thats how it works and thanks for points
Explanation:
<h2><u>Steps </u><u>:</u></h2>
- <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
- <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
- <u>That's</u><u> </u><u>all </u>
<u>hope</u><u> it</u><u> </u><u>help</u>
<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
Explanation:
The given data is as follows.
Length (l) = 2.4 m
Frequency (f) = 567 Hz
Formula to calculate the speed of a transverse wave is as follows.
f = 
Putting the gicven values into the above formula as follows.
f = 
567 Hz = 
v = 544.32 m/s
Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.