Answer:
0.4113772 s
Explanation:
Given the following :
Mass of bullet (m1) = 8g = 0.008kg
Initial horizontal Velocity (u1) = 280m/s
Mass of block (m2) = 0.992kg
Maxumum distance (x) = 15cm = 0.15m
Recall;
Period (T) = 2π√(m/k)
According to the law of conservation of momentum : (inelastic Collison)
m1 * u1 = (m1 + m2) * v
Where v is the final Velocity of the colliding bodies
0.008 * 280 = (0.008 + 0.992) * v
2.24 = 1 * v
v = 2.24m/s
K. E = P. E
K. E = 0.5mv^2
P.E = 0.5kx^2
0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2
0.5*1*5.0176 = 0.5*k*0.0225
2.5088 = 0.01125k
k = 2.5088 / 0.01125
k = 223.00444 N/m
Therefore,
Period (T) = 2π√(m/k)
T = 2π√(0.992+0.008) / 233.0444
T = 2π√0.0042910
T = 2π * 0.0655059
T = 0.4113772 s
Answer:
1472.98 m
Explanation:
Data provided:
Speed of circular looping, v = 340 m/s
Acceleration, a = 8g
here,
g is the acceleration due to the gravity = 9.81 m/s²
Now,
the centripetal acceleration is given as,
r is the radius of the loop
on substituting the respective values, we get
or
r = 1472.98 m
Answer:
Momentum is 100 kg.m/s
Explanation:
given
mass, m = 5 kg
velocity, v = 20 m/s
To find : momentum (P)
We know that momentum is given by equation:
p = mv
= 5 kg x 20 m/s
= 100 kg.m/s
Answer:
mu=12Tm^2
Explanation:
the magnetic moment mu of a single loop is given by:
where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:
hope this helps!!
