The Moon s escape speed will be smaller than Earth's.
The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.
The formula of escape speed is
where
v is escape velocity
G is universal gravitational constant
M is mass of the body to be escaped from
r is distance from the center of the mass
we can say that,
Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.
we know that,
The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.
since, v ∝ g
The Moon s escape speed will be smaller than Earth's.
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Answer:
The particle which completes the given equation is :
Explanation:
The given reaction is of a fission reaction:
Total mass on the reactant side is equal to the total mass on the product side:
239 + 1 = 100 +A+ 2
A = 138
Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:
94 + 1(0) = 40 + Z + 2(0)
Z = 54
So atomic number 54 id of Xenon.
The particle which completes the given equation is :
Answer:
Explanation:
If the initial velocity is U
Then the horizontal component of the velocity is
Ux= Ucosθ
Then the range for a projectile is give as
R=Ux.t
Where t is the time of flight
The time of flight is given as
t=2USinθ/g
Therefore,
R=Ux.t
R=UCosθ.2USinθ/g
R=U^2×2SinθCosθ/g
Then, from trigonometric ratio
2SinθCosθ= Sin2θ
R=U^2Sin2θ/g
Given that θ=32° and g=9.81m/s^2
Then
R=U^2Sin2×32/9.81
R=U^2Sin64/9.81
R=0.0916U^2
Then, range is given by R=0.0916U^2
A=0.0916U^2.
T
The box is at a distance A from the point of projection. Then the range R=A
R=0.0916U^2
A=0.0916U^2
Then,
U^2=A/0.0916
U^2=10.915A
Then the initial velocity should be
U=√10.915A
U=3.3√A
direction ................................................................
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>
