Answer:
im pretty sure its b
Note that when the acceleration is negative — on the interval [0, 2) — that means that the velocity is decreasing. When the acceleration is positive — on the interval (2, 4] — the velocity is increasing. Speeding up and slowing down.
Answer:
Explanation:
A )
Th expression for time period for a spring -mass system is as follows
T = 2π
So if mass is doubled , Time period will be √2 or 1.414 times or 5.9 s
B )
If the mass is halved , time period becomes 1 / √2 times or .70 times or 3.0 s
C)
Time period does not depend upon the amplitude of oscillation . So in this case time period will be unchanged ie 4.2 s
D )
As per the formula above , if spring constant k is doubled , time period will be 1 / √2 times or .70 times or 3.0 s
Answer:
a. d₁/d₂ = 1.09 b. 0.054 mW
Explanation:
a. What is the ratio of the diameter of the first student's eardrum to that of the second student?
We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²
So, I ∝ I/d²
I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.
Given that I₂ = 1.18I₁
I₂/I₁ = 1.18
Since I₁/I₂ = d₂²/d₁²
√(I₁/I₂) = d₂/d₁
d₁/d₂ = √(I₂/I₁)
d₁/d₂ = √1.18
d₁/d₂ = 1.09
So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09
b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?
We know intensity, I = P/A where P = acoustic power and A = area = πd²/4
Now, P = IA
= I₂A₂
= I₂πd₂²/4
= 1.18I₁πd₂²/4
Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m
So, P = 1.18I₁πd₂²/4
= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4
= 0.691244π × 10⁻⁴ W/4 =
2.172 × 10⁻⁴ W/4
= 0.543 × 10⁻⁴ W
= 0.0543 × 10⁻³ W
= 0.0543 mW
≅ 0.054 mW
Between B and C, the object was going at a constant velocity; it is going 60 m/min consistently for that time frame.
Answer:
Work done, W = 10195.92 Joules
Explanation:
Given that,
Mass of the crate, m = 170 kg
Distance, d = 10.2 m
The coefficient of friction, 
Let W is the work done by the mover. It is given by in terms of coefficient of friction as :


W = 10195.92 Joules
So, the work done by the mover is 10195.92 Joules. Hence, this is the required solution.