Answer:
Injections of aqueous solution of fruct
levulose of formula C, H, O,,
For
prevent dehydration such solutions are obtained
dissolving a mass m = 25g of fructose for a volume of 50
final solution.
1.1 Calculate the molecular molar mass of fructose
1.2 Determine the amount of corresponding fructose material
1.3 Calculate the molar concentration of these fructose solutions
1.4
Explanation:
It is given that,
Mass of an electron, 
Initial speed of the electron, 
Final speed of the electron, 
Distance, d = 5 cm = 0.05 m
(a) The acceleration of the electron is calculated using the third equation of motion as :



Force exerted on the electron is given by :



(b) Let W is the weight of the electron. It can be calculated as :



Comparison,


Hence, this is the required solution.
Answer:
a) The strength of gravity decreases if one moved away from Jupiter
b) The strength of gravity increases if one fell into Jupiter
Explanation:
The gravitational attraction is given by Newton law of gravitation as follows;

Where;
G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
M = The mass of Jupiter
m = The mass of the nearby body
R = The distance between the centers of Jupiter and the body
From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies
Therefore, as one moves away, R increases, and the strength of gravity reduces
Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.
Snapping a leaf shut around an insect, I think.
Answer:
2/3
Explanation:
In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.
In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:
Rocket speed = rocket momentum / rocket mass.
As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:
Initial state: Speed = rocket momentum / rocket mass.
Final state: Speed = 2 rocket momentum / 3 rocket mass. -------------> 2/3