Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
![x - x_o = u_xt](https://tex.z-dn.net/?f=x%20-%20x_o%20%3D%20u_xt)
---- (1)
the equation of the motion y is :
![\mathtt{y - y_o =u_yt - 0.5 gt^2}](https://tex.z-dn.net/?f=%5Cmathtt%7By%20-%20y_o%20%3Du_yt%20-%200.5%20gt%5E2%7D)
![\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}](https://tex.z-dn.net/?f=%5Cmathtt%7By%20%3D%20u_yt-4.9t%5E2%20%20%20%20%20%5C%20%5C%20%5C%20%20since%20%28y_o%20%3D0%29%7D)
![\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }](https://tex.z-dn.net/?f=%5Cmathtt%7B%201%3D%20%28u%20%5C%20sin%2040%5E0%29t%20-4.9%20%5C%20t%5E2%20%20%20%20%20%20%20%20%7D)
![\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}](https://tex.z-dn.net/?f=%5Cmathtt%7B1%20%3D%208%20sin%2040%5E0%20t%20-%204.9%20t%5E2%7D)
![\mathtt{1 = 5.14t - 4.9t^2}](https://tex.z-dn.net/?f=%5Cmathtt%7B1%20%3D%205.14t%20-%204.9t%5E2%7D)
![\mathtt{4.9t^2 - 5.14t +1 = 0}](https://tex.z-dn.net/?f=%5Cmathtt%7B4.9t%5E2%20-%205.14t%20%2B1%20%3D%200%7D)
By using the quadratic formula, we have;
![\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }](https://tex.z-dn.net/?f=%5Cmathtt%7B%20%5Cdfrac%7B%20-b%20%5Cpm%20%5Csqrt%7Bb%5E2%20-%204ac%7D%7D%7B2a%7D%7D%20%20%20%20%20%7D)
where;
a = 4.9, b = -5.14 c = 1
![= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%20-%28-5.14%29%20%5Cpm%20%5Csqrt%7B%28-5.14%29%5E2%20-%204%284.9%29%281%29%7D%7D%7B2%284.9%29%7D%7D%20%20%20%20%20%7D)
![= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%205.14%20%5Cpm%20%5Csqrt%7B26.4196%20-19.6%7D%7D%7B9.8%7D%7D%20%20%20%20%20%7D)
![= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%205.14%20%5Cpm%20%5Csqrt%7B6.8196%7D%7D%7B9.8%7D%7D%20%20%20%20%20%7D)
![= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%205.14%2B%20%5Csqrt%7B6.8196%7D%7D%7B9.8%7D%20%20%5C%20%20%5C%20OR%20%5C%20%20%5C%20%20%5Cdfrac%7B%205.14-%20%5Csqrt%7B6.8196%7D%7D%7B9.8%7D%7D%20%20%20%20%7D)
![= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%205.14%2B%202.6114%7D%7B9.8%7D%20%20%5C%20%20%5C%20OR%20%5C%20%20%5C%20%20%5Cdfrac%7B%205.14-%202.6114%7D%7B9.8%7D%7D%20%20%20%20%7D)
![= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }](https://tex.z-dn.net/?f=%3D%20%5Cmathtt%7B%20%5Cdfrac%7B%207.7514%7D%7B9.8%7D%20%20%5C%20%20%5C%20OR%20%5C%20%20%5C%20%20%5Cdfrac%7B%202.5286%7D%7B9.8%7D%7D%20%20%20%20%7D)
![= \mathbf{ 0.791 \ \ OR \ \ 0.258} }](https://tex.z-dn.net/?f=%3D%20%5Cmathbf%7B%200.791%20%5C%20%20%5C%20OR%20%5C%20%20%5C%20%200.258%7D%20%20%20%20%7D)
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
![\mathtt{x = u_x(0.258)}](https://tex.z-dn.net/?f=%5Cmathtt%7Bx%20%3D%20u_x%280.258%29%7D)
![\mathtt{x = ucos 40^0 (0.258)}](https://tex.z-dn.net/?f=%5Cmathtt%7Bx%20%3D%20ucos%2040%5E0%20%280.258%29%7D)
![\mathtt{x = 8 \ cos 40^0 (0.258)}](https://tex.z-dn.net/?f=%5Cmathtt%7Bx%20%3D%208%20%5C%20cos%2040%5E0%20%280.258%29%7D)
![\mathbf{x = 1.581 \ m}](https://tex.z-dn.net/?f=%5Cmathbf%7Bx%20%3D%201.581%20%20%5C%20m%7D)
Thus, the child is 1.581 m far from the fence
A baseball traveling at 100 mph has more kinetic energy than a baseball traveling at 50 mph because the kinetic energy = 1/2 x mass x velocity. Since the baseballs should have the same mass, the velocity is what will determine which ball has more kinetic energy. Since the 100 mph baseball has a higher velocity than the 50 mph baseball, it has more kinetic energy.
Types of Gas? I'm not exactly sure what you're asking
Answer:
The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.
shdh
Answer:
The current would be same in both situation.
Explanation:
Given that,
Current I = 13 A
Number of turns = 23
We need to calculate the induced emf
Using formula of induced emf is
![\epsilon=NA\dfrac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3DNA%5Cdfrac%7BdB%7D%7Bdt%7D)
For N = 1
![\epsilon=A\dfrac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3DA%5Cdfrac%7BdB%7D%7Bdt%7D)
We need to calculate the current
Using formula of current
![i=\dfrac{\epsilon}{R}](https://tex.z-dn.net/?f=i%3D%5Cdfrac%7B%5Cepsilon%7D%7BR%7D)
Put the value of emf
![i=\dfrac{A\dfrac{dB}{dt}}{R}](https://tex.z-dn.net/?f=i%3D%5Cdfrac%7BA%5Cdfrac%7BdB%7D%7Bdt%7D%7D%7BR%7D)
Now, if the number of turn is 22 , then induced emf would be
![\epsilon'=NA\dfrac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%27%3DNA%5Cdfrac%7BdB%7D%7Bdt%7D)
Then the current would be
![i'=\dfrac{\epsilon'}{NR}](https://tex.z-dn.net/?f=i%27%3D%5Cdfrac%7B%5Cepsilon%27%7D%7BNR%7D)
![i'=\dfrac{NA\dfrac{dB}{dt}}{NR}](https://tex.z-dn.net/?f=i%27%3D%5Cdfrac%7BNA%5Cdfrac%7BdB%7D%7Bdt%7D%7D%7BNR%7D)
![i'=\dfrac{A\dfrac{dB}{dt}}{R}](https://tex.z-dn.net/?f=i%27%3D%5Cdfrac%7BA%5Cdfrac%7BdB%7D%7Bdt%7D%7D%7BR%7D)
![i'=i](https://tex.z-dn.net/?f=i%27%3Di)
Hence, The current would be same in both situation.