Answer:
0.43622 seconds
0.9158 foot
-4.43 ffot/sec
Explanation:
we first find the period of oscillation
= 2π√w/gk
= 2π√4000/32x2600
= 2π√0.00481
= 2π0.0694
= 0.43622
b. we find the angular velocity
2π/T
= 2π/0.43622
= 14.41 rad/sec
we find displacement
rom the calculation in the attachment
Ф = -144.1
initial condition
1*cos(-144.1 rad)
= 0.9158 foot
initial velocity of the car
= (-1)(14.41)sin(ω(0)-144.1)
= -4.44 foot/sec
Answer:
Density of 127 I =
Also,
Explanation:
Given, the radius of a nucleus is given as
.
where,
The density of the nucleus is defined as the mass of the nucleus M per unit volume V.
For the nucleus 127 I,
Mass, M =
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by
On comparing with the density of the solid iodine,
The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
