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:
A law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
Answer:
The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Explanation:
Given that,
Mass = 2 kg
Radius = 0.5 m
Angular speed = 3 rad/s
Force = 10 N
(I). We need to calculate the rotational kinetic energy
Using formula of kinetic energy
(II). We need to calculate the instantaneous change rate of the kinetic energy
Using formula of kinetic energy
On differentiating
....(I)
Using newton's second law
Put the value of a in equation (I)
Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
According to the conservation of energy
- Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed
Mass is 2kg=m
#A
h=5m
PE
Now
- KE=98J
- 1/2mv²=98J
- 1/2×2v²=98J
- v²=98J
- v=√98
- v=9.4m/s
#B
h=3.2m
PE:-
Now
- KE=62.7J
- 1/2mv²=62.7
- v²=62.7
- v=√62.7
- v=7.9m/s
#C
h=2m
PE
Now