Answer:
I believe it's frictional force
________b_____ 7 km east
|
| 2km north.
|a
|
°
pythagorean theorem : ✓a² + b² = c²
c² = a² + b² = 4 + 49 = 53
c = ✓53 km
displacement = c = ✓53 km
distance = 10 + 3 + 2 = 15 km
Answer:
Given that
m = 5.3 kg
Fx = 2x + 4
We know that work done by force F given as
w= ∫ F. dx
a)
Given that x=1.08 m to x=6.5 m
Fx = 2x + 4
w= ∫ F. dx
![w=\left [x^2+4x \right ]_{1.08}^{6.5}](https://tex.z-dn.net/?f=w%3D%5Cleft%20%5Bx%5E2%2B4x%20%5Cright%20%5D_%7B1.08%7D%5E%7B6.5%7D)
w=62.7 J
b)
We know that potential energy given as
∫ dU = -∫F.dx ( w= ∫ F. dx)
ΔU= -62.7 J
c)
We know that form work power energy theorem
Net work = Change in kinetic energy
W= KE₂ - KE₁
62.7 =KE₂ - (1/2)x 5.3 x 3²
KE₂ = 86.55 J
This is the kinetic energy at 6.5m
Answer:
Explanation:
distance travelled
s = 2πR
= 2 X 3.14 X 140
= 880 m
final velocity = v
initial velocity = u
distance travelled = s
time = 60 s
s = ut + 1/2 at²
880 = .5 x a x 60²
a = .244 m/s²
final velocity v = at
= .244 x 60
= 14.66
centripetal acceleration at final moment
v² /R
14.66 X 14.66 / 140
= 1.53 m/s⁻²
1.53 m/s²
this is centripetal acceleration which acts towards the centre.
tangential acceleration calculated a _t = .244
redial acceleration ( centripetal ) = 1.53
Resultant acceleration
R²= 1.53² + .244 ²
R = 1.55 m/s²
total force = 1.55 x 76
= 118 N
Concave mirrors can make a real image.when the object is away from the mirror it is called real image and when the object is closer to the mirror it is called virtual mirror.
