To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are
Comet mass 
Radius 
Rock was dropped from a height 'h' from surface = 1m
The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is
Where G means gravitational universal constant and M the mass of the planet
Now calculate the value of the time
The time taken for the rock to reach the surface is t = 87.58s
Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)
The big bang produced dark energy, which accounts for some of the energy needed to expand the universe.
The vastness of space may contain a type of matter known as “dark matter.”
The universe is currently expanding at an accelerating rate.
Hope this helps !
The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:
where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
Learn more about vector addition:
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brainly.com/question/5892298
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Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped:
![M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]](https://tex.z-dn.net/?f=M-m%20%3D%20%5Cfrac%7BF%7D%7Bg-a%7D%20-%20%5Cfrac%7BF%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B1%7D%7Bg-a%7D%20-%20%5Cfrac%7B1%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%28g%2B%28a%2F2%29%29%20-%20%28g-a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7Bg%2B%28a%2F2%29%20-%20g%20%2B%20a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%283a%2F2%29%7D%7Bg%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7Bag%7D%7B2%7D%7D%20%5D)
