Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer:
The dust present in the clouds.
Explanation:
The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.
It is exactly this dust that protects the molecules against the action of ultraviolet rays.
We will convert the 1dm3 in terms of cm3 as follows:
1dm^3 = (10 cm)^3
= 1000 cm^3
The mass of platinum is equal to 900 lb.
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g
Then density of platinum is equal to 21.4 g/cm^3
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3
= 19076.6 cm^3
The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.
Answer:
the correct answer is A, the object goes 4 times as far
Explanation:
This is a projectile launching approach. Where the parameter we are controlling is the initial speed and they ask us how far it goes from the initial one. Let's calculate the range with a speed (vo)
R1 = v₀² sin 2θ / g
Now let's double vo, the new speed is
v = 2 v₀
We calculate the scope
R2 = (2v₀)² sin 2θ / g
R2 = 4 v₀² sin 2θ / g
R2 = 4 R1
Therefore the correct answer is A, the object goes 4 times further
Explanation:
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