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,

Learn more about angular motion here:
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This phenomena is also called the Doppler shift. When the source of light is approaching towards an observer, the color tends to be blue shifted, but when the source is moving away or being stretch, the color tends to red shifted. In astronomy it can be use how fast galaxy is moving towards us or how fast it moves away.
Answer:
Can't see anything, please share clearly
Answer:
Impulse = 80Ns
Explanation:
Given the following data;
Mass = 3kg
Force = 20N
Time = 4 seconds
To find the impulse experienced by the object;
Impulse = force * time
Impulse = 20*4
Impulse = 80Ns
Therefore, the impulse experienced by the object is 80 Newton-seconds.
Answer:
966 mph
Explanation:
Using as convention:
- East --> positive x-direction
- North --> Positive y-direction
The x- and y- components of the initial velocity of the jet can be written as

While the components of the velocity of the wind are

So the components of the resultant velocity of the jet are

And the new speed is the magnitude of the resultant velocity:
