Answer:
Explanation:
Given
Mass = m
Time = t
Force = mass × acceleration ..... 1
Where acceleration = velocity/time
Ave Force = m×∆v/∆t
= mv/t
Answer:
6.78 km
Explanation:
Length of path due east = 4.3km
Length of path south = 2.48km
Unknown:
Distance covered = ?
Solution:
The distance covered is the total length of path from start to finish. It takes cognizance of the turns and every direction moved.
Unlike displacement which only considers the net direction from start to finish, distance sums up the total path.
So;
Distance = 4.3km + 2.48km = 6.78km
Answer:
![v = 4.375\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%204.375%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:
![(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}](https://tex.z-dn.net/?f=%28350%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%29%5Ccdot%20%281.5%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20-%20%282%5C%2Cm%29%5Ccdot%20%2860%5C%2Ckg%29%5Ccdot%20v%20%3D%200%5C%2Ckg%5Ccdot%20%5Cfrac%7Bm%5E%7B2%7D%7D%7Bs%7D)
The initial speed of Ryan is:
![v = 4.375\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%204.375%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
b) a = -k / m x
, c) d²x / dt² = - A w² cos (wt+Ф)
, d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt
+Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
Answer:
Upright, Reduced
Explanation:
Given that,
Height of the object, h = 20 cm
Image distance, v = 25 cm
Height of the image, h' = 5 cm
We know that the magnification of the mirror is given by :
![m=\dfrac{h'}{h}=\dfrac{v}{u}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7Bh%27%7D%7Bh%7D%3D%5Cdfrac%7Bv%7D%7Bu%7D)
![\dfrac{5}{20}=\dfrac{25}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B20%7D%3D%5Cdfrac%7B25%7D%7Bu%7D)
u = -100 cm (negative due to sign conventions)
Let f is the focal length of the lens. It can be calculated as :
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D)
![\dfrac{1}{f}=\dfrac{1}{25}-\dfrac{1}{(-100)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7B25%7D-%5Cdfrac%7B1%7D%7B%28-100%29%7D)
f = 20 cm
It is seen that the focal length of the lens is positive. So, it is a convex lens. The image formed by the convex lens is Upright, Reduced. Hence, the correct option is (a).