Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
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Answer:
0.0984
Explanation:
The first diagram below illustrates a free body diagram that will aid in resolving this problem.
According to the diagram, the force's horizontal component can be expressed as:
Substituting 42° for θ and 87.0° for
Meanwhile, the vertical component is:
Again substituting 42° for θ and 87.0° for
In resolving the vector, let A denote the components in mutually perpendicular directions.
The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ
The frictional force can be expressed as:
Where;
is the coefficient of friction
N = the normal force
Also, the normal reaction (N) is calculated as mg - F sin θ
Substituting . Normal reaction becomes:
By balancing the forces, the horizontal component of the force equals the frictional force.
The horizontal component is described as follows:
Rearranging the equation above to isolate leads to:
Substituting in the following values:
m = 73 kg
g = 9.8 m/s²
Thus:
Therefore, the coefficient of friction is = 0.0984
Answer:
a)n= 3.125 x electrones.
b)J= 1.515 x A/m²
c) =1.114 x
m/s
d) ver explicación
Explanation:
La corriente 'I' = 5A =>5C/s
diámetro 'd'= 2.05 x m
radio 'r' = d/2 => 1.025 x m
número de electrones 'n'= 8.5 x
a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
Como sabemos que: Q= ne
donde e es la carga del electrón, es decir, 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrones.
b) La densidad de corriente 'J' en el cable se calcula como
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) La velocidad típica '' de un electrón se expresa como:
=
=1.515 x / 8.5 x
x |-1.6 x
|
=1.114 x
m/s
d) De acuerdo con estas ecuaciones,
J= I/A
=
=
Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?
Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².