Answer: 37.981 m/s
Explanation:
This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:
<u>x-component:
</u>
(1)
Where:
is the point where the ball strikes ground horizontally
is the ball's initial speed
because we are told the ball is thrown horizontally
is the time since the ball is thrown until it hits the ground
<u>y-component:
</u>
(2)
Where:
is the initial height of the ball
is the final height of the ball (when it finally hits the ground)
is the acceleration due gravity
Knowing this, let's start by finding
from (2):
<u></u>
(3)
(4)
(5)
(6)
Then, we have to substitute (6) in (1):
(7)
And find
:
(8)
(9)
(10)
On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity
:
(11)
(12)
(13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.
However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).
Answer:
5.6ft
Explanation:
on avgerage a male gets 5 foot 6 inches
The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be
C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
![f=\frac{v_T}{4L}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bv_T%7D%7B4L%7D)
where, ![v_T=v\sqrt{\frac{T}{273} }](https://tex.z-dn.net/?f=v_T%3Dv%5Csqrt%7B%5Cfrac%7BT%7D%7B273%7D%20%7D)
Since, T=37+273 K = 310 K
v = 331 m/s
![\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s](https://tex.z-dn.net/?f=%5Ctherefore%20v_T%3D331%5Csqrt%7B%5Cfrac%7B310%7D%7B273%7D%20%7D%20%3D%20352.72%20%5C%20m%2Fs)
Using this, we get:
![f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B352.72%7D%7B4%280.240%29%7D%20%5C%5Cf%3D367.42%20%5C%20Hz)
Hence, the fundamental frequency is 367.42 Hz.
To learn more about Attention here:
brainly.com/question/14673613
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Answer:
atoms cannot go bad
Explanation:
Because they stay alive and get good nutriants
Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = ![X_{L} = j\omega L = 2\pi fL](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20j%5Comega%20L%20%3D%202%5Cpi%20fL)
where
R = resistance
![X_{L} = Inductive Reactance](https://tex.z-dn.net/?f=X_%7BL%7D%20%3D%20Inductive%20Reactance)
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
![C = \frac{\epsilon_{o}A}{x}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7DA%7D%7Bx%7D)
where
x = separation between the parallel plates
Thus
C ∝ ![\frac{1}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D)
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
![Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28X_%7BL%7D%20-%20X_%7BC%7D%29%5E%7B2%7D%7D)
Also,
Z ∝ I
Therefore,
![\frac{Z}{I} = \frac{Z'}{I'}](https://tex.z-dn.net/?f=%5Cfrac%7BZ%7D%7BI%7D%20%3D%20%5Cfrac%7BZ%27%7D%7BI%27%7D)
![\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%7D%7B3I%7D%20%3D%20%5Cfrac%7B%5Csqrt%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%7D%7BI%7D)
![{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20%28R%20-%20X_%7BC%7D%29%5E%7B2%7D%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20%28R%20-%20%5Cfrac%7BX_%7BC%7D%7D%7B3%7D%29%5E%7B2%7D%7D%29)
![{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})](https://tex.z-dn.net/?f=%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20X_%7BC%7D%5E%7B2%7D%20-%202RX_%7BC%7D%20%3D%209%28%7BR%5E%7B2%7D%20%2B%20R%5E%7B2%7D%20%2B%20%5Cfrac%7BX_%7BC%7D%5E%7B2%7D%7D%7B9%7D%20-%202RX_%7BC%7D%29)
Solving the above eqn:
![X_{C} = 4R](https://tex.z-dn.net/?f=X_%7BC%7D%20%3D%204R)