The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
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Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=
y-component of resultant vector=
x-component of equilibrium vector=
y-component of equilibrium vector=
Because equilibrium vector lies in III quadrant
The angle lies in III quadrant
In III quadrant ,angle =
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
Answer:
1331.84 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement = 490 km
a = Acceleration
g = Acceleration due to gravity = 1.81 m/s² = a
From equation of linear motion
The speed of the material must be 1331.84 m/s in order to reach the height of 490 km
The current flowing in silicon bar is 2.02 10^-12 A.
<u>Explanation:</u>
Length of silicon bar, l = 10 μm = 0.001 cm
Free electron density, Ne = 104 cm^3
Hole density, Nh = 1016 cm^3
μn = 1200 cm^2 / V s
μр = 500 cm^2 / V s
The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.
J = Je + Jh
J = n qE μn + p qE μp
where, n and p are electron and hole densities.
J = Eq (n μn + p μp)
we know that E = V / l
So, J = (V / l) q (n μn + p μp)
J = (1.6 10^-19) / 0.001 (104
1200 + 1016
500)
J = 1012480 10^-16 A / m^2.
or
J = 1.01 10^-9 A / m^2
Current, I = JA
A is the area of bar, A = 20 μm = 0.002 cm
I = 1.01 10^-9
0.002 = 2.02
10^-12
So, the current flowing in silicon bar is 2.02 10^-12 A.