acceleration = Velocity changes ÷ time of the velocity changes
4 m/s^2 =
4 × 10^(-3) × 3600 km / h =
4 × 3.6 =
14.4 km / h
Thus :
14.4 = V(2) - V(1) / t(2) - t(1)
14.4 = V(2) - 20 / 10
Multiply both sides by 10
10 × 14.4 = 10 × ( V(2) - 20 ) / 10
144 = V(2) - 20
Add both sides 20
144 + 20 = V(2) - 20 + 20
V(2) = 164 Km/h
Thus the final velocity after 10 seconds is 164 Km/h .
im not gonna write a research paper but this is the really easy way write global warming talk about animals the polar ice caps and water levels then for what causes it burning fossil fuels and energy plants. then finish off with its awful and we should use solar or geothermic or wind or when the time comes fusion not fission fusion makes helium from hydrogen then burylliam from helium then oxygen and silicon so on so forth instead of fissions uranium,plutonium and thorium and with radioactive waste
The final velocity of the bullet+block is 0.799 m/s
Explanation:
We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.
Mathematically, we can write:
where
m = 0.001 kg is the mass of the bullet
u = 800 m/s is the initial velocity of the bullet
M = 1 kg is the mass of the block
U = 0 is the initial velocity of the block (initially at rest)
v is the final combined velocity of the bullet and the block
Solving the equation for v, we find the final velocity:
Learn more about conservation of momentum:
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#LearnwithBrainly
Answer:
a) Batteries and fuel cells are examples of galvanic cell
b) Ag-cathode and Zn-anode
c) Cell notation: Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
Explanation:
a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.
A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.
b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.
In the given case, the E⁰ values from the standard reduction potential table are:
E⁰(Zn/Zn2+) = -0.763 V
E°(Ag/Ag+)=+0.799 V
Therefore, Ag will be the cathode and Zn will be the anode
c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.
Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
