consider the motion of the mass parallel to the incline
v₀ = initial velocity at the bottom of incline = 0 m/s
v = final velocity at the top of incline = 8.00 m/s
a = acceleration
d = displacement = L = length of incline = 15 m
using the equation
v² = v²₀ + 2 a d
8² = 0² + 2 a (15)
64 = 30 a
a = 64/30
a = 2.13 m/s²
F = applied force
from the force diagram, perpendicular to incline , force equation is given as
N = mg Cos30
μ = Coefficient of friction = 0.426
frictional force acting on the mass is given as
f = μ N
f = μ mg Cos30
parallel to incline , force equation is given as
F - f - mg Sin30 = ma
F - μ mg Cos30 - mg Sin30 = ma
inserting the values
F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)
F = 425.82 N
Answer:
7.5 × 10^3 m/s^2
Explanation:
use the formula that does not have v in it to solve for acceleration.
Answer:
θ₁ = 35.32°
Explanation:
given,
refractive index of medium 1 = n₁ = 1.75
refractive index of medium 2 = n₂ = 1.24
condition to describe the refracted angle
...(1)
Using Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
θ₁ , θ₂ is the angle of incidence and refractive index
n₁. n₂ is the refractive index medium 1 and medium 2
1.75 x sin θ₁ = 1.24 x sin θ₂
From equation (1)
1.75 x sin θ₁ = 1.24 x sin (90-θ₁)
1.75 sin θ₁ = 1.24 cos θ₁
tan θ₁ = 0.708
θ₁ = 35.32°
Hence, angle of incidence is equal to θ₁ = 35.32°
Answers:
a) 
b) 
c) 
d) 46000 s
Explanation:
<h2>a) Time for one cycle of the radio wave</h2>
We know the maser radiowave has a frequency 
of 
In addition we know there is an inverse relation between frequency and time 
:
(1)
Isolating : 
(2)
(3)
(4) This is the time for 1 cycle
<h2>
b) Cycles that occur in 1 h</h2>
If 
and we already know the amount of cycles per second , then:
This is the number of cycles in an hour
<h2>c) How many cycles would have occurred during the age of the earth, which is estimated to be
?</h2>
Firstly, we have to convert this from years to seconds:
Now we have to multiply this value for the frequency of the maser radiowave:
This is the number of cycles in the age of the Earth
<h2>
d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>
If we have 1 second out for every 100,000 years, then:
This means the maser would be 46000 s off after a time interval equal to the age of the earth
