Answer:
O. 5 g
Explanation:
It literally says for every cm3 its 0.5 g
Answer:
(a) 20 m
(b) 6 m/s²
(c) Between t=0 and t=2, the body moves to the left.
Between t=2 and t=4, the body moves to the right.
Explanation:
v = 3t² − 6t
x(0) = 4
(a) Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 6t) dt
x = t³ − 3t² + C
Use initial condition to find value of C.
4 = 0³ − 3(0)² + C
4 = C
x = t³ − 3t² + 4
Find position at t = 4.
x = 4³ − 3(4)² + 4
x = 20
(b) Acceleration is the derivative of velocity.
a = dv/dt
a = 6t − 6
Find acceleration at t = 2.
a = 6(2) − 6
a = 6
(c) v = 3t² − 6t
v = 3t (t − 2)
The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.
When 0 < t < 2, v < 0.
When t > 2, v > 0.
Answer:
It is producing either a 435-Hz sound or a 441-Hz sound.
Explanation:
When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:
(1)
In this problem, we know that:
- The frequency of the first trombone is 
- 6 beats are heard every 2 seconds, so the beat frequency is
If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:
Answer:
6.37 x 10⁷Hz
Explanation:
Given parameters:
Wavelength = 4.71m
Unknown:
Frequency = ?
Solution:
To solve this problem, we use the expression below:
V = F∧
V is the velocity = 3 x 10⁸m/s
F is the frequency
∧ is the wavelength
3 x 10⁸ = F x 4.71
F = 63694267.52Hz = 6.37 x 10⁷Hz
