Answer:
Gravity
Explanation:
The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
<span>You are given a submerged submarine accelerating upward at 0.325 m/s</span>² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.
Amazingly awesome !
Go Luke !
See Luke investigate !
Water, water, water.
Heat, heat, heat.
Luke rocks !
He probably learned something.
For a better understanding of the question, please see attached picture that I've created.
We need to identify the other two angles by applying the Law of sine where we are given with the following values:
a = 3.2 unit
b = 2.4 unit
c = 4.6 unit
∠A = unknown
∠B = unknown
∠C = 110°
Solving for ∠B, we have:
sin C / c = sin B / b
sin 110 / 4.6 = sin B / 2.4
sin B = 2.4*sin 110 / 4.6
B = 29.36°
Solving for ∠A, we have:
sinC/c = sinA/ a
sin110 / 4.6 = sinA / 3.2
sinA = 3.2*sin110/ 4.6
A = 40.82°
Therefore, the missing angles are ∠A=40.82° and ∠B=29.36°.