Answer:
A) 17.7 m/s
B) 15.98 m
C) Zero
E) 9.8 m/s²
Explanation:
given information
distance, h = - 34 m
time, t = 5 s
A) What is the initial speed of the egg?
h - h₀ = v₀t - 
t², h₀ = 0
- 34 = v₀ 5 - \frac{1}{2} 9.8 5²
- 34 = 5 v₀ - 122.5
v₀ = 122.5 - 34/5
= 17.7 m/s
B) How high does it rise above its starting point?
v² = v₀² - 2gh
v = 0 (highest point)
2gh = v₀²
h = v₀²/2g
= 17.7²/2 (9.8)
= 15.98 m
C) What is the magnitude of its velocity at the highest point?
v = 0 (at highest point)
E) What are the magnitude and direction of its acceleration at the highest point?
g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.
Speed = wavelength × frequency
speed = 10/1000 × 5.0
speed = 0.001 × 5.0
speed = 0.005m/s
KE = ½m*v² = ½*1.0*[0.866*3E8]² = 3.375E16 J
<span>Etot = mc²/√[1 - (v/c)²] = 1.8E17 J</span>
<span>1.37 m/s
Assuming her initial velocity is totally horizontal and her vertical velocity is only affected by gravity, let's first calculate how much time she has until she reaches the ledge 8.00 m below her.
d = 1/2AT^2
8.00m = 1/2 * 9.8 m/s^2 * T^2
Solve for T
8.00 m = 4.9 m/s^2 * T^2
Divide both sides by 4.9 m/s^2
1.632653061 s^2 = T^2
Take square root of both sides
1.277753 s = T
So we now know that she has 1.277753 seconds in which to reach a horizontal distance of 1.75 m. So how fast does she need to be going?
1.75 m / 1.277753 s = 1.369592 m/s
Since we only have 3 significant figures in our data, round the result to 3 figures giving 1.37 m/s</span>
Answer:
Spring constant = YA / L
Explanation:
Let F be the force being applied on cross sectional area A of metal bar due to which an extension of ΔL is obtained in the wire of length L then
stress = F / A
strain = ΔL /L
Young's modulus = ( F / A ) / (ΔL /L)
Y = ( F L / A ΔL )
( F / ΔL ) = ( YA / L )
Spring constant = ( F / ΔL )
Spring constant = YA / L
