Answer:
I think the correct answer is the skier at the top.
Explanation:
cause potential energy is equal to mgh.
here mass and gravitational acceleration is same for all 3, but the top fellow has more height
By V=IR
A: 24=I*20
I = 1.2A
B: 220 = I*250
I = 0.88A
C: 6= I*3
I = 2 A
C,A,B
A simple electric has two terminals i.e. negative and positive, And it has a top and a bottom namely Metal Cap and Metal disc.
<h3><u>Description of each part :-</u></h3>
- <u>Metal</u><u> </u><u>cap</u><u>:</u><u>-</u><u> </u><em>The</em><em> </em><em>part</em><em> </em><em>which</em><em> </em><em>is</em><em> </em><em>above</em><em> </em><em>the</em><em> </em><em>positive</em><em> </em><em>terminal</em><em>,</em><em> </em><em>a</em><em> </em><em>cap</em><em> </em><em>of</em><em> </em><em>metal</em><em> </em><em>through</em><em> </em><em>which</em><em> </em><em>the</em><em> </em><em>positively</em><em> </em><em>charged</em><em> </em><em>electrons</em><em> </em><em>are</em><em> </em><em>passed</em><em>.</em>
- <u>Metal</u><u> </u><u>disc</u><u>:</u><u>-</u><u> </u><em>The part which is </em><em>below</em><em> the </em><em>negative</em><em> terminal, a </em><em>disc</em><em> of metal through which the </em><em>negatively</em><em> </em><em>charged</em><em> electrons are passed.</em>
- <u>Positive</u><u> </u><u>terminal</u><u>:</u><u>-</u><u> </u><em>In</em><em> </em><em>a</em><em> </em><em>circuit</em><em> </em><em>the</em><em> </em><em>current</em><em> </em><em>flows</em><em> </em><em>from</em><em> </em><em>positive</em><em> </em><em>terminal</em><em> </em><em>to</em><em> </em><em>negative</em><em> </em><em>terminal</em><em>.</em><em> </em>
- These terminals are also known as cathode and anode.
- Cathode is the other name for positive terminal and anode is the other name for negative terminal.
Answer:
v = 47.85 m / s
, θ = 64.7º
Explanation:
This is a missile throwing exercise.
Let's find the speed to reach the maximum height, at this point the vertical speed is zero
= v_{oy}^{2} - 2 g y
0 = v_{oy}^{2} - 2gy
v_{oy} = √2gy
let's calculate
v_{oy} = √ (2 9.8 21.3)
v_{oy} = 20.43 m / s
now we can calculate the time it takes to get to this point
vy = v_{oy} - g t
t = v_{oy} / g
t = 20.43 / 9.8
t = 2.08 s
in projectile launching, the time it takes for the body to rise is the same as the time it takes to go down, so the total launch time is
= 2 t
t_{v} = 2 2.08 = 4.16 s
let's use the horizontal throw ratio
x = v₀ₓ t_{v}
v₀ₓ = x / t_{v}
v₀ₓ = 180 / 4.16
v₀ₓ = 43.27 m / s
initial velocity is
v = √ (v₀ₓ² + v_{oy}^{2})
v = √ (20.43² + 43.27²)
v = 47.85 m / s
with an angle of
tan θ = I go / vox
θ = tan⁻¹ (43.27 / 20.43)
θ = 64.7º
Answer:
In the double-entry system, transactions are recorded in terms of debits and credits. Since a debit in one account offsets a credit in another, the sum of all debits must equal the sum of all credits.
Explanation: