The types of meteorites are iron, stone and iron-stony
meteorites. The stony-iron meteorites account for less than 2% of all known
meteorites. They comprised of roughly equal amounts of nickel-iron and stone
and are divided into two groups: pallasites and mesosiderites. The answers are
only B and C
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
Answer:
Because they have already made an impact within our atmosphere
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
Increase; newton’s second law of motion
