Answer:
time taken = 7.514 seconds
Here given:
using the given formula:
How much negative charge has been removed from a positively charged electroscope if it has a charge of 7.5x10-11C?
If C is Coulombs then, −1 C is equivalent to the charge of approximately 6.242×10^18 electrons.
You already gave the answer: -7.5x10–11C
It is like asking: What colour is my white horse?
Now if you want to know how many electrons have been removed, then multiplying the number of electrons in a coulomb by the coulombs you have, you get 4.6815 x 10^8 electrons. Which is almost the same answer as before.
Answer:
Protons, Electrons, and Neutrons are the 3 primary particles in an atom.
Protons - (+1)
Electrons - (-1)
Neutrons - (0)
<h3>Your answer would be C</h3>
Answer:
d) Law of Conservation of Energy
Explanation:
The first law of thermodynamics states that energy can neither be created nor destroyed; energy can only change from one form to another.
For multiple choice you can use an elimination method. If you remember the law then you know it's not A or C because "energy cannot be created nor destroyed", and it's not B because it has nothing to do with mass.
Hope this helps!
Answer:
v₂ = - 1.29 m / s
, K₁ / Em = 0.996
, K₂ / Em = 0.004
Explanation:
This is an exercise for the moment, where the system is formed by the shotgun, the man arm and the bullet, in this case the forces are internal therefore the moment is conserved
Initial. Before shooting
p₀ = 0
Final. After shooting
= m v₁ + M v₂
Where index 1 is for the bullet and index 2 for the shotgun, arm and man set
The mass of the bullet is m = 0.04 kg
The mass of the set M = 3.0 + 12.5 = 15.5 kg
p₀ =
0 = m v₁ + M v₂
v₂ = - m / M v₁
v₂ = - 0.04 / 15.5 400
v₂ = - 1.29 m / s
The mechanical energy is equal to the kinetic energy, let's calculate the energy for each of the two elements
K₁ = ½ m v₁²
K₁ = 1/2 0.04 400²
K₁ = 3200 J
K₂ = ½ M v₂²
K₂ = ½ 15.5 1.29²
K₂ = 12.90 J
The total energy of the system is the sum of the energy of each component
Em = K₁ + K₂
Em = 3200 + 12.90
Em = 3212.9 J
The fraction of transmitted energy is
To the bullet
K₁ / Em = 3200 / 3212.9
K₁ / Em = 0.996
To the system
K₂ / Em = 12.9 / 3212.9
K₂ / Em = 0.004