Answer:
my pleasure
Explanation:
5.6kg = 5600 grams
multiply the mass value by 1000
Answer:
Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.
Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm
Answer:
This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.
Explanation:
σ=E*ε
ε=δ/L
σ=E*δ/L
δ=(σ*L)/E
σ=F/A
δ=(F*L)/(A*E)
As Force,Area and Length is same
δ∞1/E
From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.
Answer:
i = 0.3326 L
Explanation:
A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is
2 L = n λ n = 1, 2, 3…
The first harmonic or leather for n = 1
Wave speed is related to wavelength and frequency
v = λ f
λ = v / f
Let's replace in the first equation
2 L = 1 (v / f₁)
For the shortest length L = L-l
2 (L- l) = 1 (v / f₂)
These two equations form our equation system, let's eliminate v
v = 2L f₁
v = 2 (L-l) f₂
2L f₁ = 2 (L-l) f₂
L- l = L f₁ / f₂
l = L - L f₁ / f₂
l = L (1- f₁ / f₂)
.
Let's calculate
l / L = (1- 309/463)
i / L = 0.3326
