Answer:
∝ = 28.92 rad/s²
Explanation:
Applying third equation of motion to the angular motion, we have:
2∝θ = ωf² - ωi²
where,
∝ = angular acceleration = ?
θ = angular displacement = (27 rev)(2π rad/1 rev) = 169.64 rad/s
ωf = final angular velocity = 99 rad/s
ωi = initial angular velocity = 0 rad/s
Therefore,
(2)∝(169.64 rad) = (99 rad/s)² - (0 rad/s²)
∝ = (9801 rad²/s²)/(38.8 rad)
<u>∝ = 28.92 rad/s²</u>
Answer:
tension is 37.8 N
Explanation:
given data
mass of bar m1 = 2 kg
length of bar L = 1.4 m
suspended mass m2 = 5 kg
suspended object position length L2 = 0.8 m
to find out
tension
solution
we consider here bar is connected with hinge and
we know here system is equilibrium
so here net torque will be zero at joint
and mass 2 kg act at L1 = 1.4 /2 = 0.7 m
so torque = m1×g× ( L1 ) + m2 ×g× (L2) - T(L)
so
2 ×9.8 × ( 1.4/2) + 5×9.8 × ( 0.8) - T(1.4) = 0
T = 52.98 / 1.4
T = 37.8
so tension is 37.8 N
Answer:
Explanation:
given,
charge of two spherical drop = 0.1 nC
potential at the surface = 300 V
two drops merge to form a single drop
potential at the surface of new drop = ?
r = 0.003 m
volume =
=
= 2.612 × 10⁻⁷ m³
R = 0.00396 m
The total energy remains the same, as long as
none of it escapes the closed system.
<u>Answer:</u>
The velocity is 30.279 m/s
<u>Explanation</u>:
Consider the initial speed of the semi-trailer be v
Then, initial kinetic energy =
According to question, the semi-trailer coast along a ramp, which is inclined at an angle of 170, and to a distance of 160m to stop
Change in vertical position == 46.779m
Final potential energy of semitrailer = mgh
Applying principle of conservation of energy,
= mgh
Solving for v, we get = 2gh = 2*9.8*46.779 = 916.8684
= 916.8684
v = 30.279 m/s
Therefore, the velocity is 30.279 m/s