Answer:
9.934 m/s²
Explanation:
Given:
Initial speed of the Bugatti Veyron Super Sport = 0 mi/h
Final speed of the Bugatti Veyron Super Sport = 60 mi/h
Now,
1 mi/h = 0.44704 m / s
thus,
60 mi/h = 0.44704 × 60 = 26.8224 m/s
Time = 2.70 m/s
Now,
The acceleration (a) is given as:
thus,
or
a = 9.934 m/s²
Answer:
Explanation:
As we know that the wave equation is given as
now we have
so we have
also we have
so we have
now we know that at t = 0 and x = 0 wave is at y = 0.19 m
so we have
so we have
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)
Answer:
to move an object from its place or position
Explanation:
if you move an object from where it first was
Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v = 
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c
