consider the motion of bullet as it penetrates through the block
v₀ = initial velocity of the bullet = 400 m/s
v = final velocity of the bullet = 0 m/s
d = distance traveled = 12 cm = 0.12 m
a = acceleration = ?
Using the equation
v² = v₀² + 2 a d
0² = 400² + 2a (0.12)
a = - 6.67 x 10⁵ m/s²
m = mass of the bullet = 10 g = 0.010 kg
The resistive force is given as
f = ma
inserting the values
f = (0.010) (- 6.67 x 10⁵ )
f = - 6670 N
b)
t = time taken by the bullet
using the equation
v = v₀ + a t
0 = 400 + (- 6.67 x 10⁵ ) t
t = 6 x 10⁻⁴ sec
c)
Explanation:
It is given that,
Initial vapor pressure, P₁ = 77.86 mm
Initial temperature, T₁ = 318.3 K
Final vapor pressure, P₂ = 161.3 mm
Initial temperature, T₂ = 340.7 K
We need to find its heat of vaporization. It can be calculated by using Clausius-Clapeyron equation.
So, the heat of vaporization of a substance is 29.31 kJ/mol. Hence, this is the required solution.
Answer:
<h2>index of refraction = 1.69</h2><h2>percentage error = 10.58%</h2>Explanation:
According to Snell's law, the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. The constant is known as the refractive index.
Mathematically
i = angle of incidence measured = 63.5°
r = angle of refraction measured = 32°
n = refractive index
The index fraction calculated is approx. 1.69.
If the index of refraction of a clear plastic as listed in the book is 1.89 and the calculated is 1.69, the percentage error will be calculated as thus;
%error =
%error =
%error =
%error = 10.58%
Answer:
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
The damping constant can be found by
Finally, the mass m can be found by
Where g is approximately 32 ft/s²
Therefore, the required differential equation is
The initial position is
The initial velocity is
Answer:
-252.52
Explanation:
L = Distance between lenses = 10 cm
D = Near point = 25 cm
= Focal length of objective = 0.9 cm
= Focal length of eyepiece = 1.1 cm
Magnification of a compound microscope is given by
The angular magnification of the compound microscope is -252.52