Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q = 
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q = 
q = 3.6 10⁵ C
Answer:
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg
Explanation:
Hi there!
Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:
PE = EPE
m · g · h = 1/2 k · x²
Where:
m = mass.
g = acceleration due to gravity.
h = height.
k = spring constant.
x = compression distance
The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:
EPE =1/2 k · x²
EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J
Then, using the equation of gravitational potential energy:
PE = m · g · h = 541.2 J
m = 541.2 J/ g · h
m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)
m = 16.6 kg
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.
What are you asking here?
Answer:
0.0675 seconds
Explanation:
From the question,
We apply newton's second law of motion
F = m(v-u)/t.................... Equation 1
Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.
make t the subject of the equation
t = m(v-u)/F................... Equation 2
Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)
Substitute these value into equation 2
t = 180(0-6.0)/-1600
t = -1080/-1600
t = 0.0675 seconds.
Answer:
5p
Explanation:
We are given that a tin has 50 electrons.
We have to find in which subshell electrons experience the lowest effective nuclear charge.
We know that the electron in outermost shell experience the lowest effective nuclear charge.
Electronic configuration is given by
Outer most sub-shell is 5p. Therefore, 5p subshell experience the lowest effective nuclear charge because the distance of 5p sub-shell is large from nucleus.
Answer: 5p
