The velocity of the tip of the second hand is 0.0158 m/s
Explanation:
First of all, we need to calculate the angular velocity of the second hand.
We know that the second hand completes one full circle in
T = 60 seconds
Therefore, its angular velocity is:
Now we can calculate the velocity of a point on the tip of the hand by using the formula
where
is the angular velocity
r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)
Substituting,
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Answer:
Option C. 5,000 kg m/s
Explanation:
<u>Linear Momentum on a System of Particles
</u>
Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed
P=mv
The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object
The sum of the momenta of both objects prior to the collision is
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Explanation:
It is given that,
Force, 
Position vector, 
(a) The torque on the particle about the origin is given by :
(b) To find the angle between r and F use dot product formula as :
Hence, this is the required solution.
