Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as
Distance = speed x time
From the information given,
speed = 320 m/s
time = 4.5 s
By substituting these values into the formula, we have
Distance = 320 m/s x 4.5s
s cancels out. We are left with m. Thus,
Distance = 1440m
Answer:
0.06 Kg
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Net Force (F) = 3 N
Mass (m) =?
Next, we shall determine the acceleration of the object. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Acceleration (a) =?
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Divide both side by 0.18
a = 9 / 0.18
a = 50 m/s²
Finally, we shall determine the mass of the object. This can be obtained as follow:
Net Force (F) = 3 N
Acceleration (a) = 50 N
Mass (m) =?
F = ma
3 = m × 50
Divide both side by 50
m = 3 / 50
m = 0.06 Kg
Therefore, the mass of the object is 0.06 Kg
The alpha particle is emitted at 4235 m/s
Explanation:
We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:
where:
is the mass of the original nucleus
is the initial velocity of the nucleus
is the mass of the alpha particle
is the final velocity of the alpha particle
is the mass of the daughter nucleus
is the final velocity of the nucleus
Solving for , we find the final velocity of the alpha particle:
Learn more about momentum:
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Answer:
18.03 N
Explanation:
From the fiqure below,
Using parallelogram law of vector
R² = 15²+5²-2×5×15cos(180-60)
R² = 225+25-150cos120°
R² = 250-150(-0.5)
R² = 250+75
R² = 325
R = √325
R = 18.03 N
Hence the resultant force of the object is 18.03 N
