Answer:
Current, I = 0.0011 A
Explanation:
It is given that,
Diameter of rod, d = 2.56 cm
Radius of rod, r = 1.28 cm = 0.0128 m
The resistivity of the pure silicon, 
Length of rod, l = 20 cm = 0.2 m
Voltage, 
The resistivity of the rod is given by :


R = 893692.30 ohms
Current flowing in the rod is calculated using Ohm's law as :
V = I R


I = 0.0011 A
So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
Answer:
C. 
Explanation:
0 charge → <em>Neutron</em>
1 charge → <em>Proton</em>
I am joyous to assist you anytime.
Answer:
A.c
Explanation:
The chromosphere is above the photosphere, the visible "surface" of the Sun. It lies below the solar corona, the Sun's upper atmosphere, which extends many thousands of kilometers above the chromosphere into space. The plasma (electrically charged gas) in the chromosphere has a very low density.
In basic terms it is the 2nd one out from the core.
Answer:
A. No
B. si
Explanation:
A. El trabajo realizado en la carga es la energía potencial ganada por la carga al elevar la carga al nivel del camión y colocar la carga dentro del camión.
El trabajo realizado para elevar la carga W = m × g × h
Dónde;
m = masa de la carga
g = aceleración debido a la gravedad
h = Nivel de altura donde se coloca la carga en el camión
Por lo tanto, el trabajo realizado depende de la masa, m, de la carga y el nivel de altura, h, donde la carga se coloca en el camión y el trabajo realizado es el mismo para todos los métodos utilizados para colocar la carga en el camión
B. La ecuación para el trabajo realizado, W, también se puede escribir de la siguiente manera;
W = Fuerza, F × Distancia, D
De lo que tenemos;
F = W/D
Por lo tanto, cuando la mesa aumenta la distancia, como una rampa o un plano inclinado, la fuerza requerida disminuirá.