The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.
<h3>Final velocity of the block/bullet system</h3>
Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.
K.E = P.E
¹/₂mv² = mgh
¹/₂v² = gh
v² = 2gh
v = √2gh
where;
- h is the maximum height reached by the system
- v is the initial velocity of the system
v = √(2 x 9.8 x 1.1)
v = 4.64 m/s
<h3>Initial velocity of the bullet</h3>
Apply the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
- u₁ is the initial velocity of the bullet
- u₂ is the initial velocity of the block
- v is the final velocity after collision
- m₁ is mass bullet
- m₂ is mass of block
(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)
0.0075u₁ = 4.4428
u₁ = 4.4428/0.0075
u₁ = 592.37 m/s
<h3>Initial kinetic energy of the bullet</h3>
K.Ei = ¹/₂m₁u₁²
K.Ei = ¹/₂(0.0075)(592.37)²
K.Ei = 1,315.88 J
<h3>Final kinetic energy of the block/bullet system</h3>
K.Ef = ¹/₂(m₁ + m₂)v²
K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²
K.Ef = 10.31 J
<h3>Ratio of final kinetic energy to initial kinetic energy</h3>
= K.Ef/K.Ei x 100%
= (10.31 / 1,315.88) x 100%
= 0.78 %
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Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
F = normal force by each board on each side
W = weight of the board in between acting in down direction = 95.5 N
f = frictional force in upward direction by each board
= coefficient of friction = 0.663
Using equilibrium of force in Upward direction
f + f = W
f = W/2
f = 95.5/2 = 47.75 N
frictional force is given as
f = F
47.75 = (0.663) F
F = 72.02 N
Answer:
b) G.P.E = Mgh
300j = M x 10 m/s² x 15 m
300 j/ 10 m/s² x 15 m = M
300j/ 150 s² = M
2kg = M
c) K.E = 1/2 m v²
K.E = 1/2 (50) (50)²
K.E = 1/2 (50) (2500)
K.E= 125000/2
K.E = 625 000 J
Answer:
B.It is a satellite that collects data about rain and snow
C.Its orbit covers 90 percent of Earth’s surface
F.The sensors measure microwaves
