Precisely around 1,800 miles below.
We're told that the planets have EQUAL MASS.
If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.
Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.
'A' says this.
'B' is totally absurd, because it talks about gravity repelling things.
'C' says exactly the opposite for the two planets.
'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.
Explanation:
Equation for Impact
FΔt = ΔP,
F = force
Δt = Impact of time
ΔP = Change in momentum
Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.
Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.
Road safety containers are loaded with liquid or sand as they improve the period of impact.
Answer:
a) vB = 10.77 ft/s
b) W = 11.30 lb*ft
Explanation:
a) W = 8 lb ⇒ m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug
vA <em>lin</em> = 5 ft/s
rA = 2 ft
v <em>rad</em> = 4 ft/s
vB = ?
rB = 1 ft
W = ?
We can apply The law of conservation of angular momentum
L<em>in</em> = L<em>fin</em>
m*vA*rA = m*vB*rB ⇒ vB = vA*rA / rB
⇒ vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s (tangential speed)
then we get
vB = √(vB tang² + vB rad²) ⇒ vB = √((10 ft/s)² + (4 ft/s)²)
⇒ vB = 10.77 ft/s
b) W = ΔK = K<em>B</em> - K<em>A</em> = 0.5*m*vB² - 0.5*m*vA²
⇒ W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)
⇒ W = 11.30 lb*ft
Answer:
C.
m
Explanation:
We are given that
Weight of board=w=10 N
Length of board=L=5 m
Tension in the string=T=3 N
Applied upward force=F=7 N
We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.
Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.
The board is uniform therefore, the center of board is the mid- point of board.
Therefore, the lever arm of weight=
Now, the torque exerted by the weight of the board
The torque exerted by applied force=
In static equilibrium
The sum of rotational forces=0
The two rotational force act in opposite direction therefore,
Substitute the values
m
Hence, option C is true.
