Answer:
F = 4.147 × 10^23
v = 1.31 × 10^4
Explanation:
Given the following :
mass of Jupiter (m1) = 1.9 × 10^27
Mass of sun (m2) = 1.99 × 10^30
Distance between sun and jupiter (r) = 7.8 × 10^11m
Gravitational force (F) :
(Gm1m2) / r^2
Where ; G = 6.673×10^-11 ( Gravitational constant)
F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2
F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)
F = (25.231 × 10^46) / (60.84 × 10^22)
F = 3.235 × 10^(46 - 22)
F = 0.4147 × 10^24
F = 4.147 × 10^23
Speed of Jupiter (v) :
v = √(Fr) / m1
v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)
v = √32.3466 × 10^(23+11) / 1.9 × 10^27
v = √32.3466× 10^34 / 1.9 × 10^27
v = √17. 023 × 10^34-27
v = √17.023 × 10^7
v = 13047.221
v = 1.31 × 10^4
There is no "why", because that's not what happens. The truth is
exactly the opposite.
Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.
The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air. So the force that lifts the
object in water is greater than the force that lifts it in air, and the object
appears to weigh less in the water.
Answer:
a) 
b) x = 4.47 cm
c) 
d) x = 1.48 cm
Explanation:
a) The center of mass is equal to:
Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃
b) If
m₁ = 23g
m₂ = 15 g
m₃ = 58 g
d₁ = 1.1 cm
d₂ = 1.9 cm
d₃ = 3.2 cm
c) The center of the mass of the beads realtive to the center of bead is:
d) 
Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Answer: The correct answer is A). Animal burrow because burrow fossils represent the preserved byproducts of behavior rather than physical remains, they are considered a kind of trace fossil. One common kind of burrow fossil is known as Skolithos, and the similar Trypanites, Ophiomorpha and Diplocraterion.
