Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm = 
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
C
Explanation:
got a one hundred on the test
Answer:
please give me brainlist and follow
Explanation:
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA = 
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
