We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
They are incline hope this helps!
Answer:
The circumference of the Earth is 24818.58 miles
Explanation:
Analysis conceptual : The formula of the circumference is the following:
L= π*D Formula (1)
Where:
L : is the length of the circumference in miles (mi)
π : is the constant
D : is the diameter of the circumference in miles (mi)
Known data
π = 3.1416
D= 7900 miles: Diameter of the Earth
Problem development
We apply the formula 1 to calculate the circumference of the Earth (L):
L= π*7900 miles
L= 24818.58 miles
Atoms like carbon and nitrogen do not form ions because the electronegativity of these atoms are not that high nor very low which means electrons are fairly stable in the atom. While chlorine has very high electronegativity and for sodium very low, atoms tend to receive or release electrons.
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
