Answer:
Option (B)
Explanation:
L = 0.5 H, I = 2 A, R = 10 Ohm
V = I R
In case of dc the inductor cannot works.
V = 2 x 10 = 20 V
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Answer:
Explanation:
Let the velocity of firing be u at angle θ
At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ
So , vtop = v cosθ
At height h/2
vertical component of velocity v₂
v₂² = (usinθ)² - 2 g . h/2
v₂² = u²sin²θ - gh
horizontal component of velocity at height h/2 = u cosθ
velocity at height h / 2
= √ ( u²sin²θ - gh + u² cos²θ)
Given
√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop
u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ
u²sin²θ - gh = 3 u² cos²θ
At height h , vertical component of velocity is zero
0 = u²sin²θ - 2gh
gh = u²sin²θ / 2
u²sin²θ - u²sin²θ / 2 = 3 u² cos²θ
u²sin²θ / 2 = 3 u² cos²θ
Tan²θ = 6
Tanθ = 2.45
θ = 68⁰ .
Answer:
Explanation:
Given that,
Mass of the object, m = 29 kg
Torsion constant of the wire, K = 1.14 N-m
Number of cycles, n = 98
Time, t = 74 s
To find,
The rotational inertia of the object.
Solution,
Relationship between the moment of inertia, time period and the torsion constant of the spring is given by :
Where I is the moment of inertia
K is spring constant
Let T Is the time period of oscillation, such that,
So, the rotational inertia of the object is .
Answer:
Explanation:
It is given that,
Mass of puck 1,
Mass of puck 2,
Initial speed of puck 1,
Initial speed of puck 2,
After the collision, the speed of puck 1,
Let is the final velocity (in m/s) of puck 2 after the collision. Using the conservation of momentum as :
So, the final velocity of the puck 2 after the collision is 24 m/s. Hence, this is the required solution.