Complete question:
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.
Answer:
The bottom current is 12.8 A to the right.
Explanation:
Given;
length of the wires, L = 3.0 m
current in the top wire, I₁ = 12.5 A
repulsive force between the two wires, F = 2.4 x 10⁻⁴ N
distance between the two wires, r = 40 cm = 0.4 m
The repulsive force between the two wires is given by;
Where;
I₂ is the bottom current
The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.
Therefore, the bottom current is 12.8 A to the right.
Answer:
4 s
Explanation:
Given:
Δx = 12 m
v₀ = 6 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
12 m = ½ (0 m/s + 6 m/s) t
t = 4 s
Answer:
The block has an acceleration of 
Explanation:
By means of Newton's second law it can be determine the acceleration of the block.
(1)
Where 
represents the net force, m is the mass and a is the acceleration.
(2)
The forces present in x are 
and 
(the friction force):
Notice that 
subtracts to 
since it is at the opposite direction.
The forces present in y balance each other:
Therefore:
(3)
But 
and writing (3) in terms of a it is get:
So the block has an acceleration of .
Answer:
i) Distancia, ii) La cinta métrica es impracticable.
Explanation:
i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).
La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.
ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.
En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.
Answer:
a.) a = 0 ms⁻²
b.) a = 9.58 ms⁻²
c.) a = 7.67 ms⁻²
Explanation:
a.)
Acceleration (a) is defined as the time rate of change of velocity
Given data
Final velocity = v₂ = 0 m/s
Initial velocity = v ₁ = 0 m/s
As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,
a = 0 ms⁻²
b.)
Given data
As the space shuttle start from rest, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 8 min = 480 s
By the definition of Acceleration (a)
a = 9.58 ms⁻²
c.)
Given data
As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero
Initial velocity = v₁ = 0 ms⁻¹
Final velocity = v₂ = 4600 ms⁻¹
Time = t = 10 min = 600 s
By the definition of Acceleration (a)
a = 7.67 ms⁻²
