Answer:
v2 = 27.3m/s
Explanation:
Assuming forward as positive.
Mass = m1 = 64kg
Let v be the common velocity of the student and the skateboard.
mass of skateboard = m2 = 5.94kg
v = 1.4m/s
Since the skateboard and the student are initially moving together at the same velocity their momentum together is
(m1 + m2)v
Let the final velocity of the student be v1 and the final velocity of the skateboard be v2
v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)
Then from conservation of momentum, momentum before is equal to momentum after.
(m1 + m2)v = m1v1 + m2v2
m2v2= (m1 + m2)v – m1v1
v2 = ( (m1 + m2)v – m1v1)/m2
v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94
v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94
v2 = 27.3m/s
Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
where v is final velocity which is zero at max height and u is it initial
hence
now we can find time in the 15 cm ascent
using quadratic formula
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
t=0.0409
Answer:
the answer is the spinning of the moon lets us see different amounts of light
Explanation:
you wanna know why uh yes ok lets cut to the magic so when the moon.
Answer:
It is an SI unit
Explanation:
The metre is defined as the length of the path travelled by light in a vacuum in 1299 792 458 of a second. The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole
