Answer:
It is frequently stated that the value of the acceleration due to gravity at the pole is larger than at the equator because the poles are closer to the center of the earth due to the earth's oblateness. ... The measured value is larger because the earth's density is not uniform but increases toward the center.
Answer:
skateboard b
Explanation:
p=mv
skateboard a
p=(60kg)(1.5m/s)=90kg*m/s
skateboard b
p=(50kg)(2m/s)=100kg*m/s
Answer:
ΔT = 0.02412 s
Explanation:
We will simply calculate the time for both the waves to travel through rail distance.
FOR THE TRAVELING THROUGH RAIL:

FOR THE WAVE TRAVELING THROUGH AIR:

The separation in time between two pulses can now be given as follows:

<u>ΔT = 0.02412 s</u>
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment 
Mass of the heavier fragment 
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum













Hence, The speed of the heavier fragment is 0.335c.