Answer
given,
mass of student = 60 Kg
mass of suitcase = 10 Kg
a) Work done by picking of Suitcase is equal to zero
b) acceleration = 0.1 m/s²
distance = 2 m
Using second law conservation
F = m a
F = 10 x 0.1 = 1 N
c) Work done
W = F x s
W = 1 x 2 = 2 J
d) When the are moving with constant speed acceleration is equal to zero
F = m a
F = 10 x 0 = 0 N
W = F x s = 0 x s = 0 J
e) work done = change in kinetic energy
K.E = 2 J
Sedimentary rocks are the result of igneous rocks<span> breaking down.This is a slow process so there is not as much broken down as there is not broken down.</span>
Answer:
a) The velocity of rock at 1 second, v = 9.8 m/s
b) The velocity of rock at 3 second, v = 29.4 m/s
c) The velocity of rock at 5.5 second, v = 53.9 m/s
Explanation:
Given data,
The rock is dropped from a bridge.
The initial velocity of the rock, u = 0
a) The velocity of rock at 1 second,
Using the first equation of motion
v = u + gt
v = 0 + 9.8 x 1
v = 9.8 m/s
b) The velocity of rock at 3 second,
v = u + gt
v = 0 + 9.8 x 3
v = 29.4 m/s
c) The velocity of rock at 5.5 second,
v = u + gt
v = 0 + 9.8 x 5.5
v = 53.9 m/s
Answer:
Explanation:
Given,
The angle of the slide=
The mass of the child is= m
coefficient of friction = 0.20
when she slides down now apply Newton's law
therefore the acceleration
![a=g[\sin \theta -\mu \cos \theta]](https://tex.z-dn.net/?f=a%3Dg%5B%5Csin%20%5Ctheta%20-%5Cmu%20%5Ccos%20%5Ctheta%5D)
![a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]](https://tex.z-dn.net/?f=a%3D9.8%5Ctimes%20%5B%5Csin%2042%5E%5Ccirc%20-0.2%5Ctimes%20%5Ccos%2042%5E%5Ccirc%5D)
hence, the magnitude of acceleration during her sliding is equal to
Explanation:
Take south to be negative.
a. Momentum is mass times velocity.
p = mv
p = (540 kg) (-6 m/s)
p = -3240 kg m/s
p = 3240 kg m/s south
b. Impulse = change in momentum
J = Δp
Since the mass is constant:
J = mΔv
J = (540 kg) (-4 m/s − (-6 m/s))
J = 1080 kg m/s
J = 1080 kg m/s north
