Easy ! EVERY element and every compound melts, then boils and becomes a gas, if you heat it to a high enough temperature. That includes iron, gold, water, salt, glass, almost any substance.
Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
Answer: 0.067 s
Explanation:s = Ut + 1/2at^2
0.6 = 9t + 0.5 *10 *t^2
Where a = g =10m/s/s
Solving the quadratic equation
5t^2 + 9t - 0.6=0,
t= 0.067 s and - 1.7 s
Of which 0.067 s is a valid time
Answer:
The rate of change of the area when the bottom of the ladder (denoted by
) is at 36 ft. from the wall is the following:
![\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%7C_%7Bb%3D36%7D%3D-571.2%5C%2C%20ft%5E2%2Fs)
Explanation:
The Area of the triangle is given by
where
(by using the Pythagoras' Theorem) and
is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.
The area is then
![A=\sqrt{l^2-b^2}b](https://tex.z-dn.net/?f=A%3D%5Csqrt%7Bl%5E2-b%5E2%7Db)
The rate of change of the area is given by its time derivative
![\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28%5Csqrt%7Bl%5E2-b%5E2%7D%5Ccdot%20b%5Cright%29)
![\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28%5Csqrt%7Bl%5E2-b%5E2%7D%5Cright%29%5Ccdot%20b%2B%5Cfrac%7Bdb%7D%7Bdt%7D%5Ccdot%5Csqrt%7Bl%5E2-b%5E2%7D)
Product rule
Chain rule
![\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}](https://tex.z-dn.net/?f=%5Cimplies%5Cfrac%7BdA%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B%5Csqrt%7Bl%5E2-b%5E2%7D%7D%5Ccdot%20b%5E2%5Ccdot%20%5Cfrac%7Bdb%7D%7Bdt%7D%2B%5Csqrt%7Bl%5E2-b%5E2%7D%7D%5Ccdot%20%5Cfrac%7Bdb%7D%7Bdt%7D)
![\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)](https://tex.z-dn.net/?f=%5Cimplies%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7Bdb%7D%7Bdt%7D%5Cleft%28-%5Cfrac%7B1%7D%7B%5Csqrt%7Bl%5E2-b%5E2%7D%7D%5Ccdot%20b%5E2%2B%5Csqrt%7Bl%5E2-b%5E2%7D%7D%5Cright%29)
In here we can identify
,
and
.
The result is then
![\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D8%5Cleft%28-%5Cfrac%7B1%7D%7B%5Csqrt%7B39%5E2-36%5E2%7D%7D%5Ccdot%2036%5E2%2B%5Csqrt%7B39%5E2-36%5E2%7D%7D%5Cright%29%3D-571.2%5C%2C%20ft%5E2%2Fs)
Part a)
Magnitude of electric field is given by force per unit charge
![E = \frac{F}{q}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BF%7D%7Bq%7D)
![E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B4.3%20%2A%2010%5E%7B-6%7D%7D%7B2%20%2A%2010%5E%7B-9%7D%7D)
![E = 2150 N/C](https://tex.z-dn.net/?f=E%20%3D%202150%20N%2FC)
Part b)
Electrostatic force on the proton is given as
F = qE
![F = 1.6 * 10^{-19} * 2150](https://tex.z-dn.net/?f=F%20%3D%201.6%20%2A%2010%5E%7B-19%7D%20%2A%202150)
![F = 3.44 * 10^{-16} N](https://tex.z-dn.net/?f=F%20%3D%203.44%20%2A%2010%5E%7B-16%7D%20N)
PART C)
Gravitational force is given by
![F_g = mg](https://tex.z-dn.net/?f=F_g%20%3D%20mg)
![F_g = 1.6 * 10^{-27}*9.8](https://tex.z-dn.net/?f=F_g%20%3D%201.6%20%2A%2010%5E%7B-27%7D%2A9.8)
![F_g = 1.57 * 10^{-26} N](https://tex.z-dn.net/?f=F_g%20%3D%201.57%20%2A%2010%5E%7B-26%7D%20N)
PART d)
Ratio of electric force to weight
![\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%20%5Cfrac%7B3.44%20%2A%2010%5E%7B-16%7D%7D%7B1.57%2A10%5E%7B-26%7D%7D)
![\frac{F_e}{F_g} = 2.2 * 10^{10}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%202.2%20%2A%2010%5E%7B10%7D)