The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction.
Answer:
a) 31.4 m/s
b) 50.2 m
Explanation:
a) When an object is free falling, its speed is determined by the gravity force giving it acceleration. Equation for the velocity of free fall started from the rest is:
v = g • t
g - is gravitational acceleration which is 9.81 m/s^2, sometimes rounded to 10
t - is the time of free fall
So:
v = 9.81 m/s^2 • 3.2
v = 31.4 m/s ( if g is rounded to 10, then the velocity is 10 • 3.2 = 32 m/s)
b) To determine the distance crossed in free fall we use the equation:
s = v0 + gt^2/2
v0 - is the starting velocity (since object started fall from rest, its v0 is 0)
s = gt^2/2
s = 9.81 m/s^2 • 3.2^2 / 2
s = 50.2 m (if we round g to 10 then the distance is 10 • 3.2^2/2 = 51.2 meters)
Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>
It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.
![V_x'=0.8c\\V=0.6c\\m=4*10^5kg](https://tex.z-dn.net/?f=V_x%27%3D0.8c%5C%5CV%3D0.6c%5C%5Cm%3D4%2A10%5E5kg)
- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,
![KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}](https://tex.z-dn.net/?f=KE%3Dm_0c%5E2%3D%5Cfrac%7Bmc%5E2%7D%7B1-%28%5Cfrac%7Bv_x%5E2%7D%7Bc%5E2%7D%20%29%7D)
- So, to
find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. - We have an expression from Lorents transformation for relativistic law of addition of velocities as,
![V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V](https://tex.z-dn.net/?f=V_x%27%3D%5Cfrac%7BV_x-V%7D%7B1-%5Cfrac%7BVV_x%7D%7Bc%5E2%7D%20%7D%20%5C%5Cthus%2C%5C%5CV_x%3DV_x%27%281-%5Cfrac%7BVV_x%7D%7Bc%5E2%7D%20%29%2BV)
- Substituting values, we get,
![V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c](https://tex.z-dn.net/?f=V_x%3D0.8c%281-%5Cfrac%7B0.8c%2A0.6c%7D%7Bc%5E2%7D%20%29%2B0.6c%3D%280.8c%2A0.52%29%2B0.6c%3D1.016c)
![KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B4%2A10%5E5%2A%283%2A10%5E8%29%5E2%7D%7B%5Csqrt%7B1-%5Cfrac%7B%281.016c%29%5E2%7D%7Bc%5E2%7D%20%7D%20%7D%20%3D%5Cfrac%7B1.2%2A10%5E%7B22%7D%7D%7B0.179%7D%3D6.704%2A10%5E%7B22%7DJ)
Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
Learn more about frame of reference here:
brainly.com/question/20897534
SPJ4
Answer:
it depends on wether the + and - are facing eachother
or away from eachother
Explanation: