Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer
given,
weight of the oak board = 600 N
Weight of Joe = 844 N
length of board = 4 m
Joe is standing at 1 m from left side
vertical wire is supporting at the end.
Assuming the system is in equilibrium
T₁ and T₂ be the tension at the ends of the wire
equating all the vertical force
T₁ + T₂ = 600 + 844
T₁ + T₂ = 1444...........(1)
taking moment about T₂
T₁ x 4 - 844 x 3 - 600 x 2 = 0
T₁ x 4 = 3732
T₁ = 933 N
from equation (1)
T₂ = 1444 - 933
T₂ = 511 N
During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)
