Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
1 answer:
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1. Amperes, is the SI unit (also a fundamental unit) responsible for current.
2. Δq over Δt technically
Rearrange for Δq
I x Δt = Δq
1.5mA x 5 = Δq
Δq = 0.0075
Divide this by the fundamental charge "e"
Electrons: 0.0075 / 1.60 x 10^-19
Electrons: 4.6875 x 10^16 or 4.7 x 10^16
3. So we know that the end resistances will be equal so:
ρ = RA/L
ρL = RA
ρL/A = R
Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper
We are looking for L2 so we can isolate using algebra to get:
If you fill in those values you get 0.0205
or 2.05 cm
Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s