Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
First, torque is equal to force times the distance. for the first force that is applied, the torque is zero because is applied at the hinge. so the net torque:
t = ( 12 N ) ( 0 m ) ( cos 30 ) + ( 12 N ) ( 1.68 m ) cos 45
t = 14.26 Nm is the torque with respect to the hinge
Answer:

Explanation:
As we know that backpack is kicked on the rough floor with speed "v"
So here as per force equation in vertical direction we know that

so normal force on the block is given as

now the magnitude of kinetic friction on the block is given as


now when bag is sliding on the floor then net deceleration of the block due to friction is given as


now we know that bag hits the opposite wall at L distance away in time t
so we have



Answer:
T = 0.03 Nm.
Explanation:
d = 1.5 in = 0.04 m
r = d/2 = 0.02 m
P = 56 kips = 56 x 6.89 = 386.11 MPa
σ = 42-ksi = 42 x 6.89 = 289.58 MPa
Torque = T =?
<u>Solution:</u>
σ = (P x r) / T
T = (P x r) / σ
T = (386.11 x 0.02) / 289.58
T = 0.03 Nm.
Answer:
a = 4.72 m/s²
Explanation:
given,
mass of the box (m)= 6 Kg
angle of inclination (θ) = 39°
coefficient of kinetic friction (μ) = 0.19
magnitude of acceleration = ?
box is sliding downward so,
F - f = m a
f is the friction force
m g sinθ - μ N = ma
m g sinθ - μ m g cos θ = ma
a = g sinθ - μ g cos θ
a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°
a = 4.72 m/s²
the magnitude of acceleration of the box down the slope is a = 4.72 m/s²