Answer:
500 Joules
Explanation:
W(work)= force * distance
w = 100*5
W= 500
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W=
kx² =
(530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg
=
kx² - mgx
And,
= (
kx² - mgx
)/(mg) =
]/(0.645x9.8)
= 0.78m
d) Now, if the initial initial height of block is 3
= 3 x 0.78 = 2.34m
then,
kx² - mgx - mg
=0
(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance

r is the radius of the spherical shell
dr is the thickness
volume of shell

mass of shell


now,

integrating both side



we know,




a = 9.94 m/s²
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.
The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>
Answer:
It's either B or D, I'm not positive which it is
Explanation: